Thursday

April 17, 2014

April 17, 2014

Number of results: 118,070

**Physics**

A. stopping distance = (stopping time) x (average speed) = (V/a)*(V/2) = V^2/(2a) B. If the speed doubles, with deceleration rate "a" staying the same, the stopping distance is four times farther.
*Sunday, December 27, 2009 at 7:29pm by drwls*

**Math Models**

the stopping distance D varies directly with the square of the speed S. if a car traveling 50mph has a stopping distance of 250 feet, find the stopping distance of 60mph.
*Wednesday, June 13, 2012 at 1:33pm by malekah*

**physics - stopping distance **

when a driver brings a car to a stop by breaki.g as hard as possible the stopping can be regarded as the sum of reaction distance and breaking distance. tje following data are given, v1 m/s contains 10,20,30. reaction distance are 7.5, 15, 22.5. the breaking distance are 5.0, ...
*Thursday, April 25, 2013 at 12:03pm by gladys*

**math**

If a car is speeding down a road at 40 (mph), how long is the stopping distance compared to the stopping distance if the driver were going at the posted speed limit of 25 ? Express your answer as a multiple of the stopping distance at 25
*Thursday, September 8, 2011 at 9:39am by Em*

**Physics**

This question has been answered twice already. Yes, the stopping distance quadruples if you double the speed. What is there about Stopping distance = V^2/(2a) that you don't understand?
*Monday, December 28, 2009 at 8:28pm by drwls*

**Physics - Stopping distance**

(Force)*(stopping distance) = (work done against the force) = (initial kinetic energy) Since the kinetic energies and forces are the same for the two different masses, the stopping distances are the same.
*Wednesday, October 20, 2010 at 5:56am by drwls *

**physics**

First you have to tell us what the stopping distance or stopping time is. You seem to have left that out. That will enable you to calculate the acceleration, a. Then use F = m a fr the stopping force.
*Thursday, August 2, 2012 at 1:09am by drwls*

**Physics**

You are the designer of a school bus and would like to know what the maximum stopping distance is that will ensure the students to remain in their seats. The coefficient of static friction for the passenger and the bus seat is .5, and the coefficient of kinetic friction for ...
*Sunday, November 25, 2012 at 4:55pm by Tim*

**Physics**

You are the designer of a school bus and would like to know what the maximum stopping distance is that will ensure the students to remain in their seats. The coefficient of static friction for the passenger and the bus seat is .5, and the coefficient of kinetic friction for ...
*Sunday, November 25, 2012 at 9:06pm by Tim*

**physics**

"Spotted by breaks"? Don't you mean "stopped by brakes"? Stopping distance is proportional to V^2 if the deceleration rate is the same in both cases. That assumption is usually made. The stopping distance gets multiplied by 4 when the speed doubles. Choose the first answer.
*Saturday, May 26, 2012 at 5:55am by drwls*

**physics**

on a level road the stopping distance for an automobile is 35.0m for an initail speed of 90 km/hr what is the stopping distance of the same automobile on a simalr road with the slope 1:10
*Saturday, October 30, 2010 at 4:05am by hilde*

**College physics**

According to test performed by the manufactures, an automobile with an initial speed of 75 km/h has a stopping distance of 25 m on a level road. assuming that no skidding occurs during breaking, what is the value of μ , between the wheels and the road required to achieve ...
*Tuesday, February 26, 2013 at 6:16pm by Sneakatone*

**Physics**

An automobile travelling with a speed of 60km/hr, can apply brake to stop within a distance of 20m. If the car is going twice as fast then calculate the stopping distance and stopping time? (b) Given R1 = 5.0 ± 0.2 and R2 = 10.0 ± 0.1. calculate the total resistance in ...
*Friday, July 15, 2011 at 10:07am by Shrawan*

**Statistics**

A car manufacturer is interested in conducting a study to estimate the mean stopping distance for a new type of brakes when used in a car that is traveling at 60 miles per hour. These new brakes will be installed on cars of the same model and the stopping distance will be ...
*Saturday, November 28, 2009 at 11:18pm by John*

**Statistics**

A car manufacturer is interested in conducting a study to estimate the mean stopping distance for a new type of brakes when used in a car that is traveling at 60 miles per hour. These new brakes will be installed on cars of the same model and the stopping distance will be ...
*Monday, November 30, 2009 at 10:33pm by Bob*

**physics**

When a car is traveling at 21 m/s on level ground, the stopping distance is found to be 18 m. This distance is measured by pushing hard on the brakes so that the wheels skid on the pavement. The same car is driving at the same speed down an incline that makes an angle of 7.6° ...
*Tuesday, September 25, 2012 at 6:51pm by sarah*

**physics!**

"Newton's second law in component form" is just F = m a in the horizontal direction of motion. F is the braking force, M*g*Uk where Uk is the coefficient of kinetic friction. M*g*Uk = M a means that a = g*Uk is the acceleration rate. a*(stopping time) = (initial velocity) ...
*Friday, September 23, 2011 at 8:40am by drwls*

**statistics**

im so confused! please help! 3. A car manufacturer is interested in conducting a study to estimate the mean stopping distance for a new type of brakes when used in a car that is traveling at 60 miles per hour. These new brakes will be installed on cars of the same model and ...
*Friday, March 4, 2011 at 9:19am by george*

**physics**

a 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping. find the stopping force F if the entrance velocity is 80 meter per second.
*Saturday, October 2, 2010 at 2:34am by kristel*

**Physics**

A car traveling at 20 m/s applies the brakes. The coefficients of static and kinetic friction between the tires and the road are 0.80 and 0.50 respectively. (A) Determine the stopping distance if the car is able to stop as quickly as possible without skidding. (B) Determine ...
*Thursday, October 25, 2012 at 11:09pm by Brittney*

**Physics**

1. a = F/m = 10 m/s^2 F is the friction force, m is the mass stopping time = V/a = 2.5 s stopping distance = Avg. velocity * time = 12.5 * 2.5 = 31.25 m
*Saturday, May 26, 2012 at 10:03pm by drwls*

**Physics**

a. The stopping distance is proportional to the square of the initial velocity: Ds = 2^2 * 20m = 80m = Stopping distance. The stopping time is proportional to the initial velocity: Vo = 60000m/h * (1h/3600s) = 16.67m/s. a = (Vf^2-Vo^2) / 2d, a = (0-(16.67)^2 / 40 = 6.94m/s. Ts...
*Friday, July 15, 2011 at 10:07am by Henry*

**Physics**

A 1500 kg automobile travels at a speed of 105 km/h along a straight concrete highway. Faced with an emergency situation, the driver jams on the brakes, and the car skids to a stop. (a) What will be the car's stopping distance for dry pavement (µ = 0.85)? (b) What will be the ...
*Tuesday, September 21, 2010 at 3:39am by Tom*

**Physics**

Ignore the 95.0 km/h speed of the driver. He may hit the barrier. They want you to calculate the speed V for which the stopping distance would be exactly 39.0 m. The stopping distance will be V*0.95 + (V/{a})*(V/2) = 39.0 m Solve for V. The second term, V^2/(2|a|) is the ...
*Tuesday, February 15, 2011 at 6:12pm by drwls*

**math**

predicting the stopping distance for a car traveling 45 miles per hour using this chart speed stopping (mph) Distance (FT) 55 273 56 355 75 447
*Friday, September 10, 2010 at 12:54am by Javen*

**Math**

The stopping distance d of a car in feet is related to the speed in mph by the equation d(s)=.02s^2+1.1s Calculate the speed when your stopping distance is: 1.) 51ft 2.) 315.12ft
*Saturday, February 25, 2012 at 9:24pm by Devin*

**Physics**

A race car can be slowed with a constant acceleration of -11 m/s^2 a. If the car is going 55 m/s, how many meters will it travel before it stops? b. How many meters will it take to stop a car going twice as fast? According to one of the teachers this is what they stated A. ...
*Monday, December 28, 2009 at 8:28pm by Priscilla*

**College Math**

Use the functions to find the stopping distance on wet pavement and dry pavement for a car traveling at 55 MPH.Identify the graph for each stopping distance.Explain how well your answers to item 1 model the actual stopping distances shown in Figure 3.41 on Page 384.Determine ...
*Wednesday, March 9, 2011 at 7:37am by Anonymous*

**physics**

A 1500 kg automobile travels at a speed of 105 km/h along a straight concrete highway. Faced with an emergency situation, the driver jams on the brakes, and the car skids to a stop. (a) What will be the car's stopping distance for dry pavement (µ = 0.85)? (b) What will be the ...
*Monday, September 20, 2010 at 2:11am by Tom*

**Physics**

A truck driver slams on the brakes and skids to a stop through a displacement of delta x. (a) IF the truck has twice the mass, by what factor does the stopping distance change? 1) 0.25 or 2) 1 or 3) 4 (b) If the initial velocity of the truck were halved, by what factor would ...
*Monday, February 25, 2008 at 6:47pm by John*

**Physics**

Draw a fully labelled graph of stopping distance on the indpending y,axis vs speed on the independent x,axis.plot the points and sketch your graph use the graph paper.speed 32km,48km,64km,80,km,96km,112km,stopping distance 12m,23m,36m,53m,74m,96m
*Monday, April 26, 2010 at 8:51am by Aphiwe*

**physics**

1. acceleration = -(30 m/s)/44s = -0.682 m/s^2 stopping distance = average velocity * (stopping time) = (15 m/s)*44 s = ___ 2. Time taken = (distance/(average velocity) = 70 m/13 m/s = 5.38 s Acceleration = (14 m/s)/5.38 s = 2.6 m/s^2
*Sunday, May 20, 2012 at 2:16am by drwls*

**physics - incomplete**

DON'T KNOW -- HOW MANY PASSENGERS ARE THERE? HOW FAR FROM THE STOP LIGHT DOES HE APPLY THE BRAKES? PROOFREAD YER POST BEFORE SUBMITTING IT! But you know F=ma, which will give you the acceleration v = at, which will give you the stopping time s = 1/2 at^2, which will give you ...
*Tuesday, February 7, 2012 at 12:18pm by Steve*

**math**

V1^2 / V2^2 = (40)^2 / (25)^2 = 2.56. Therefore, the stopping distance at 40 mph is 2.56 times the stopping distance at 25mph.
*Thursday, September 8, 2011 at 9:39am by Henry*

**Math help**

d=0.005s^2+0.2s where d is the stopping distance in meter and s is the speed in km per hour. use the formula to find the stopping distance when a car is travelling at 70km per hour. my answer is 38.5 is that right.
*Friday, June 18, 2010 at 9:24pm by Tim*

**math**

The stopping distance of a car traveling 25mph is 61.7ft, and for a car traveling 35mph it is 106ft. The stopping distance in feet can be described by the equation y=ax^2 + bx + c, where x is the speed in mph. Find the values of a and b.
*Friday, October 21, 2011 at 9:57pm by Alyssa*

**physics**

On level ground, the "normal force" is the weight, M*g. The force accelerating the car is M*a. It is not clear what they mean by "normal". For the stopping distance X, (stopping force)*X = initial kinetic energy
*Sunday, February 12, 2012 at 8:06am by drwls*

**Physics - Stopping distance**

Consider two bodies, A and B, moving in the same direction with the same kinetic energy. A has a mass twice that of B. If the same retarding force is applied to each, how will the stopping distances of the bodies compare?
*Wednesday, October 20, 2010 at 5:56am by Anonymous*

**Math**

The algebraic relation d= 0.0056s2 +0.14s models the relation between a vehicle's stopping distance d, in metres and its speed s, in km/h. a) What is the fastest you could drive and still be able to stop within 80m? b) What is the stopping distance for a car traveling at 120 ...
*Tuesday, January 22, 2013 at 7:15pm by Emme*

**Math**

The algebraic relation d= 0.0056s2 +0.14s models the relation between a vehicle's stopping distance d, in metres and its speed s, in km/h. a) What is the fastest you could drive and still be able to stop within 80m? b) What is the stopping distance for a car traveling at 120 ...
*Monday, January 21, 2013 at 9:45pm by Emme*

**college physics**

Assume that the bed sheet is of light material and deflects a lot when the egg hits it in the middle. Force = momentum change / time change = m delta v / delta t ---> m a The force on the egg is the mass of the egg * speed of egg when it hits / time to stop the time to stop...
*Sunday, June 1, 2008 at 2:24pm by Damon*

**Physics**

A driver approaches an intersection at a speed of 50km/h, when the light turns amber. The driver applies brakes to get the maximum stopping force without skidding. The car has a mass of 1500kg and the force of static friction between the tires and the road is 1.1*10^4N. ...
*Monday, November 12, 2012 at 5:30pm by ashley*

**physics**

a) Find the maximum rate of deceleration of a car on a dry, level road, assuming the coefficient of static friction between tires and pavement equals 1.4. (b) Find the minimum stopping distance if the car has an initial speed of 27.5 m/s. (c) Repeat parts (a) and (b) assuming ...
*Tuesday, February 8, 2011 at 10:55am by Anonymous*

**physics**

The distance that the car moves while decelerating is X = Vo^2/(2a) = 625/4 = 156.3 feet An additional distance of (25 m/s)*1 s = 25 m is travelled before the brakes are applied. The sum of the two is the stopping distance
*Tuesday, January 25, 2011 at 5:24pm by drwls*

**physics**

First of all, convert 55.1 km/h to 15.3 m/s 2a) The stopping distance X is given by this equation that relates the kinital kinetic energy to the work done against friction: (1/2)M V^2 = M*g*Uk*X Notice that the mass M cancels out, which is good since they did not tell you the ...
*Friday, December 24, 2010 at 8:54pm by drwls*

**Physics**

Add the distance travelled by the blue car while accelerating for 3.6 s to the distance travelled at constant speed for 10.1 s. You don't need the information about the yellow car or the stopping distance.
*Monday, January 24, 2011 at 11:05am by drwls*

**Algebra**

The distance required for an emergency stop for a car varies directly as the square of the car’s speed. A car traveling at 50 mph requires 140 feet to stop. What is the stopping distance for a car traveling at 30 mph? What is the stopping distance for a car traveling at 70 mph?
*Saturday, October 22, 2011 at 2:03pm by Elena*

**Physics**

The driver of a speeding empty truck (initial speed of Vi) slams on the brakes and skids to a stop through a distance d. Assuming that the brakes always supply the same stopping force, consider the following situations: 1 (a) If the truck carried a load that doubled its mass, ...
*Sunday, February 19, 2012 at 11:47pm by Raghav*

**physics**

resisting net force*distance=PE at top of board. 1110*4.1=71*g*h solve for h. height h is the total distance from the board to the stopping point.
*Wednesday, February 12, 2014 at 10:59pm by bobpursley*

**Calculus**

According to Car and Driver, an Alfa Romeo going 70 mph requires 177 feet to stop. Assuming that the stopping distance is proportional to the square of the velocity, find the stopping distance required by an Alfa Romeo going at 55 mph and at 115 mph.
*Wednesday, January 19, 2011 at 5:54pm by Brittany *

**calculus**

According to Car and Driver, an Alfa Romeo going 70 mph requires 177 feet to stop. Assuming that the stopping distance is proportional to the square of the velocity, find the stopping distance required by an Alfa Romeo going at 45 mph and at 115 mph.
*Wednesday, January 19, 2011 at 9:18pm by Jamie*

**physics**

You solve #1 and #3 by applying Newton's Second Law, F = m a. For the second problem, you first need to solve for the required deceleration rate. a = F/m = 5 m/s^2 The time it takes to stop is t = Vo/a = (30 m/s)/5 m/s^2 = 6 s The distance travelled while stopping is that ...
*Tuesday, November 11, 2008 at 10:06pm by drwls*

**physics **

First, convert speed (V) to units of m/s. 42.4 km/h = 11.78 m/s Initial kinetic energy = Work done against friction (1/2)MV^2 = M*g*Uk*X M cancels out. X = V^2/(2*g*Uk) is the stopping distance, if the car is skidding. If the brakes manage to apply the maximum force that ...
*Saturday, April 2, 2011 at 10:54pm by drwls*

**physics - stopping distance**

The reaction time is t =x/v=7.5/10 = 15/20 =22.5/30 = 0.75 s. The reaction distance for v=25 m/s is x=vt=25•0.75 = 18.75 m. The acceleration is a=v²/2s=10²/2•5=20²/2•20=30²/2•45=10 m/s² The breaking distance for v=25 m/s is s= v²/2a=25²/2•10=...
*Thursday, April 25, 2013 at 12:03pm by Elena*

**math**

Assume that stopping distance of a van varies directly with the square of the speed. A van traveling 40 miles per hour can stop in 60 feet. If the van is traveling 68 miles per hour, what is the stopping distance?
*Tuesday, August 27, 2013 at 1:51pm by misty*

**Physics**

d1=Vo*t = 28m/s * 0.8s=22.4m to react. d2 = (Vf^2 - Vo^2) / 2a, d2 = (0 - (28)^2) / -14 = 56m,stopping distance. D = d1 + d2 = 22.4 + 56 = 78.4m before stopping.
*Monday, September 19, 2011 at 8:19pm by Henry*

**physics**

No. The distance it traveled was 100m. Averagevelocity during stopping was 1/2 40 * 5
*Friday, September 7, 2007 at 5:32pm by bobpursley*

**Physics**

Sorry, I am not interested in the original speed or gravity or how long it slid across the ice. However I do need the mass of the puck. F = m *(change in velocity)/time time = distance/average speed stopping distance = .0439m average speed stopping = 23.236 m/s time = .0439/23...
*Saturday, January 8, 2011 at 5:15pm by Damon*

**Physics**

(a) Calculate accleration (a) V^2 = 2 aX. a = V^2/(2X) Then use F = ma for the force required to stop in that distance. It will be in the backwards direction (b) Add M g sin(4 degrees) to the previous answer. It is the weight component along the direction of motion. It will ...
*Thursday, September 16, 2010 at 9:06am by drwls*

**Physics**

subtract from the 35 m the distance during reactiontime (v*.9) so the stopping distance is 35-.9v. That is equal to d in ... vf^2=v^2+2ad vf is zero, solve for v.
*Tuesday, September 4, 2007 at 7:57pm by bobpursley*

**Calculus**

The total stopping distance T of a vehicle is T=2.5x+0.5x^2 where T is in feet and x is the speed in miles per hour. Approximate the change and percent change in total stopping distance as sped changes from x=25 to x=26 miles per hour. I started taking the derivative, but that...
*Tuesday, December 17, 2013 at 7:47pm by Lindsay*

**Physics**

You are travelling in your car at 50km/hr, your friend is traveling at 25 km/hr, both have the same mass. What distance do you need to travel to stop compared to the distance that your friend needs to stop? How will the stopping distance change if you and your friend are ...
*Saturday, December 8, 2012 at 1:12pm by Eric*

**Physics**

A 850.0-kg car travelling on a level road at 27.0 m/s (60.5 mi/hr) can stop, locking its wheels, in a distance of 61.0 m (200.1 ft). (a) Find the size of the horizontal force which the car applies on the road while stopping. (b) Find the stopping distance of that same car when...
*Wednesday, December 5, 2007 at 4:16pm by Lindsay*

**physics**

A car is traveling at 28 m/s when the driver spots a large pothole in the road a distance 37 m ahead. She immediately applies her brakes. If her acceleration is -7.8 m/s2, does she manage to stop before reaching the pothole? What is this stopping distance?
*Monday, September 6, 2010 at 12:32am by bill*

**physics**

A car is traveling at 51.0 mi/h on a horizontal highway If the coefficient of static friction between road and tires on a rainy day is 0.096, what is the minimum distance in which the car will stop What is the stopping distance when the surface is dry and µs = 0.603?
*Sunday, November 9, 2008 at 1:44am by nicky*

**physics**

Potential energy mgh work stopping: force*distance set them equal.
*Thursday, February 28, 2008 at 5:11pm by bobpursley*

**Physics**

A car slows to a stop over a certain distance. What could cause the stopping force to be greater?
*Tuesday, November 15, 2011 at 8:33pm by Shari*

**physics**

1. t = (Vf - Vo) / a, t = (0 - 26) / -1.5 = 17.33s. d = (Vf^2 - Vo^2) / 2a, d = (0 - (26)^2) / -3 = 225.33m = stopping distance for lead car. d = 49 + 225.33 = 274.33m = Required stopping distance for the chasing car. a = (Vf^2 - Vo^2) / 2d, a = (0 - (40)^2) / 548.66 = -2.92m/...
*Thursday, September 22, 2011 at 5:59am by Henry*

**physics**

a. V = Vo + a*t = 0 24.5 - 2*t = 0 2t = 24.5 t = 12.25 s. b. d = Vo*t + 0.5a*t^2 d = 24.5*12.25 -0.5*2*12.25^2 = 150 m. c. d = 41.6 + 150 = 191.6 m. = Required stopping distance. V^2 = Vo^2 + 2a*d a=(V^2-Vo^2)/2d=(0-29.6^2)/383.2 = -2.29 m/s^2. d. V = Vo + a*t = 0 @ stopping ...
*Thursday, January 23, 2014 at 9:13pm by Henry *

**physics**

a) initial kinetic energy = (force) x (penetration distance) Solve for the force, F = . Use kg and meters for doing the calculation. b) Vaverage * time = (Vo/2)*time = 0.15 m Solve for the stopping time. Vo is the initial velocity, 380 m/s c) a = F/m = Vo/(stopping time) (You ...
*Saturday, January 23, 2010 at 2:14am by drwls*

**physics**

A car is traveling at 45.0 mi/h on a horizontal highway. (a) If the coefficient of static friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop? ____ m (b) What is the stopping distance when the surface is dry and µs...
*Monday, November 1, 2010 at 5:45pm by natalie*

**physics hey bob**

you help me solve this problem. but how am i suppose to if i dont know the mass of the car? on a level road the stopping distance for an automobile is 35.0m for an initail speed of 90 km/hr what is the stopping distance of the same automobile on a simalr road with the slope 1...
*Sunday, October 31, 2010 at 8:53pm by anthony*

**physics**

A crate pushed along the floor with velocity \vec{v} _{i} slides a distance d after the pushing force is removed.If the mass of the crate is doubled but the initial velocity is not changed, what distance does the crate slide before stopping?
*Monday, October 22, 2012 at 11:03pm by nhan*

**physics**

Average speed x time = stopping distance The average speed during stopping is half the starting speed, or 42.5 km/h 85 km/h = 23.6 m/s 42.5 km/h = 11.8 m/s Time = 60m/11.8m/s = 5.1 seconds
*Tuesday, November 27, 2012 at 1:17pm by drwls*

**Physics**

(a) The answer is (2)(a factor of 1). The stopping distance does not change. That is because both the kinetic energy and the friction force double. (b) 4) 0.25
*Monday, February 25, 2008 at 6:47pm by drwls*

**PHYSICS**

a = Fap/m = -62,500 / 25,000 = -2.5 m/s^2. d = (Vf^2-Vo^2)/2a. d = (0-(20)^2) / -5 = 80 m. = Stopping distance.
*Thursday, February 23, 2012 at 2:51pm by Henry*

**physics**

X = Stopping distance = (reaction time)*Vo + Vo^2/(2a) The second term of that sum is the distance travelled while decelerating. Make sure Vo is in m/s. 38 km/h = 10.56 m/s X = 5.51 m + 6.56 = 12.07 m
*Monday, September 6, 2010 at 8:38pm by drwls*

**physics**

it takes 5 s to stop the car __ 55 / 11 the average speed (Va) is __ (55 + 0) / 2 the stopping distance is __ Va * 5 twice as fast means that the stopping time is doubled, along with the average speed __ this means that the car travels 4 times as far
*Wednesday, September 26, 2012 at 7:33pm by Scott*

**algebra**

Stopping distance = C r^2 270 = C*70^2 C = 270/70^2 At 50 mph, stopping distance = 50^2*270/(70^2) = (5/7)^2 * 270 ft = ?
*Wednesday, July 16, 2008 at 10:06pm by drwls*

**physics**

The cable tension will increase to T = M(g+a) when the elevator is slowing down at deceleration rate a. You can get a from the relation V^2/2 = a X where X is the stopping distance, and V is the initial velocity.
*Thursday, October 7, 2010 at 4:06pm by drwls*

**physics **

Why did you choose 9.8 as the acceleration? It is not falling. its average velocity during stopping is 89/2 km/hr (change that to m/s) So it traveled in that time a distance of avgvelocity*9s... force*distance=initial KE= 1/2 m vi^2 change vi to m/s, then solve for force.
*Thursday, October 1, 2009 at 5:06pm by bobpursley*

**Physics**

A car is traveling at 42.0 km/h on a flat highway. (a) If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop? m (b) What is the stopping distance when the surface is dry and the coefficient of ...
*Sunday, October 7, 2012 at 11:30pm by Kali*

**physics**

distance at steady speed: 1.0m/s*1.2s=1.2m stopping distance=3.4-1.2=2.2m Vf^2=Vi^2+2ad solve for a, knowing Vf=0, Vi=1m/s, d=2.2m
*Wednesday, September 21, 2011 at 11:25am by bobpursley*

**Physics **

If the force of friction is constant, prove that the stopping distance of a car on a level road varies directly with the square of the initial speed.
*Wednesday, October 6, 2010 at 6:15am by Shaila*

**physics bob**

still i cant find the normal force without mass of automabile. is normal force mass times gravity sin theta? on a level road the stopping distance for an automobile is 35.0m for an initail speed of 90 km/hr what is the stopping distance of the same automobile on a simalr road ...
*Sunday, October 31, 2010 at 9:33pm by anthony*

**physics**

A car is traveling at v = 16m/s . The driver applies the brakes and the car decelerates at a= -4.0 m/s^2 . What is the stopping distance?
*Friday, February 4, 2011 at 3:27pm by Anonymous*

**physics**

A car that weighs 1.5 × 10^4 N is initially moving at a speed of 43 km/h when the brakes are applied and the car is brought to a stop in 16 m. Assuming that the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change ...
*Friday, October 8, 2010 at 9:22pm by monnie*

**physics**

a) a car does an emergency stop from 120000m/s in 3s. what was it's deceleration? b) what was the car's average velocity and what was the stopping distance? thank you
*Wednesday, September 19, 2012 at 2:12pm by nicky*

**physics**

A car traveling at 91.0 km/h approaches the turnoff for a restaurant 30.0 m ahead. If the driver slams on the brakes with an acceleration of -6.40 m/s^2, what will her stopping distance be?
*Sunday, November 20, 2011 at 9:58pm by molly*

**Algebra**

The stopping distance D in feet for a car traveling at x miles per hour is given by d(x)= (1/12)x^2+(11/9)x. Determine the driving speeds that correspond to stopping distances between 300 and 500 feet, inclusive. Round speeds to the nearest mile per hour.
*Sunday, June 9, 2013 at 2:15pm by Kaitlyn*

**Physics**

Vo = sqrt(2 a X) Solve that for the acceleration, a. X = the stopping distance, 79 m a = Vo^2/(2X)= 4.96 m/s^2 Themn use F = M a
*Saturday, October 16, 2010 at 8:35pm by drwls*

**physics**

(1/2) m v^2 The stopping distance and time don't matter. Initial kinetic energy ends up as heat (internal energy).
*Saturday, March 23, 2013 at 5:41pm by drwls*

**algebra**

hi i was wondering if u can help me with this problem which says...stopping distance is breaking distance plus your reaction time. suppose you are told that the stoppin distance D is related to speed S by this equation: d(s)= 0.05s to the 2nd power+2.15s+1 and i need to slove ...
*Sunday, July 20, 2008 at 6:55pm by nikki*

**physics**

A car is traveling at Vx =20 m/s. The driver applies the brakes, and the car slows with ax = -4m/s^2. What is the stopping distance
*Sunday, January 29, 2012 at 10:13pm by larry*

**Physics**

A car is traveling at 24.4 m/s. If the driver has a reaction time of 1 second, what will be his stopping distance if his car slows down at 8 m/s2?
*Thursday, September 19, 2013 at 10:00pm by Joseph*

**Physics**

Determine the stopping distance for a skier moving down a slope with friction with an initial speed of 20.6m/s. Assume μk=0.178 and θ=5.27.
*Sunday, October 28, 2012 at 3:04pm by Angie*

**Pre-Calculus**

An Alfa Romeo going at 70 mph requires 169 feet to stop. Assuming that the stopping distance is proportional to the square of velocity, find the stopping distances required by an Alfa Romeo going at 35 mph and at 140 mph (its top speed).
*Sunday, February 3, 2013 at 12:28pm by Becky*

**physics**

A hockey puck sliding on the ice with velocity 10m/sec is stopped by a goalie in 0.10sec. what is the goalie stopping distance ∆x in doing so.
*Tuesday, December 7, 2010 at 12:29pm by Ama*

**Physics**

The minimum distance required to stop a car moving at 46.0 mi/h is 39.0 ft. What is the minimum stopping distance for the same car moving at 80.0 mi/h, assuming the same rate of acceleration? Can I do a proportion or is this totally different?
*Saturday, October 10, 2009 at 7:15pm by Beth*

**Physics**

Use the following equation for both: (Force)(Stopping distance) = (1/2) M V^2 V is the car velocity before collision. I assume they want you to use the same M and V for part b)
*Sunday, February 20, 2011 at 10:43pm by drwls*

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