You are driving a car 35mi/h down the road. The stoplight in front of you suddenly turns red and you slam on the brakes, skidding to a halt. You get out of the car and measure the skid marks made by your tires to be 28.5m meters. (Assume positive x to be in the forward direction).

A:What is the velocity of the car in meters per second?
B:What is the acceleration of the car?
C:If your car were going twice the velocity, how far would it take to stop if the acceleration were the same as for part (B)?
D:How long would it take the car to stop if it were going at twice the velocity?

a. convert mi/h to m/s: A nice thing to remember is that 60miles/hr =88feet/sec=26.8 m/s

b. Vf^2=vi^2+2ad solve for a.

c. You can work it out, using the same a. However, think: if you are going twice as fast, you have twice squared (4) times as much KE, so fricion distance must be four times as far.

d. average velocity is twice, distance is four times,
distance=velavb*time
time=distance/velocityavy=4/2= twice as long.

18.04

To answer these questions, we'll need to apply some basic physics formulas. Let's start with the given information:

Initial velocity (u) = 35 mi/h
Distance (s) = 28.5 m
Acceleration (a) = ?

A: What is the velocity of the car in meters per second?

To convert the velocity from miles per hour (mi/h) to meters per second (m/s), we need to use the conversion factor: 1 mi/h = 0.44704 m/s.

First, we convert the initial velocity from mi/h to m/s:

35 mi/h * 0.44704 m/s = 15.65 m/s

Therefore, the velocity of the car is 15.65 m/s.

B: What is the acceleration of the car?

To find the acceleration, we can use the formula:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s since the car comes to a halt)
u = initial velocity (15.65 m/s)
a = acceleration
s = distance (28.5 m)

Rearranging the formula, we get:

a = (v^2 - u^2) / (2s)

Substituting the values:

a = (0^2 - 15.65^2) / (2 * 28.5)
a = -245.4225 / 57
a = -4.3 m/s^2

Therefore, the acceleration of the car is approximately -4.3 m/s^2. The negative sign indicates deceleration or braking.

C: If your car were going twice the velocity, how far would it take to stop if the acceleration were the same as for part (B)?

If the car were going twice the initial velocity, it would be going at 2 * 15.65 = 31.3 m/s.

To calculate the stopping distance, we can use the same formula as in part (B):

v^2 = u^2 + 2as

Here, v = 0 m/s, u = 31.3 m/s, a = -4.3 m/s^2, and s = distance (unknown).

Rearranging the formula, we get:

s = (v^2 - u^2) / (2a)

Substituting the values:

s = (0^2 - 31.3^2) / (2 * -4.3)
s = (-978.09) / (-8.6)
s ≈ 113.8 m

Therefore, if the car were going at twice the velocity, it would take approximately 113.8 meters to come to a stop.

D: How long would it take the car to stop if it were going at twice the velocity?

To find the time taken to stop, we can use the formula:

v = u + at

Here, v = 0 m/s, u = 31.3 m/s, a = -4.3 m/s^2, and t = time taken (unknown).

Rearranging the formula, we get:

t = (v - u) / a

Substituting the values:

t = (0 - 31.3) / (-4.3)
t = 7.28 seconds (approx.)

Therefore, if the car were going at twice the velocity, it would take approximately 7.28 seconds to come to a stop.