A CAR IS MOVING INITIALLY AT A SPEED OF 50MI/H WEIGHTING 3000LB IS BROUGT TO A STOP IN A DISTANCE OF 200FT FIND THE BRAKING FORCE.AND FIND TIME REQUIRED TO STOP ASSUMING THE SAME BRAKING FORCE..AND FIND THE DISTANCE AND TIME REQUIRED TO STOP IF THE CAR WERE GOING 25MI/H INITIALLY?
Vo = 50mi/h = 22.22 m/s.
Wt. = 3000Lbs = 1362 kg.
d = 200Ft = 60.61 m.
a = -Vo^2/2d = -(22.22^2)/121.2 =
-4.07 m/s.
Fb = M*a = 1362 * (-4.07) = -5543
N. = Braking force.
V = Vo + a*t = 0.
t = -Vo/a = -22.22/-4.07 = 5.45 s. = Time required to stop.
m=1362kg,v=22.22m/s,d=61m;
we know that a moving object has a kinetic energy=1/2mv^2.So the braking action would be the negative of the work done against Kinetic energy,
W(braking)=-K.E i.e 0.5*1362*22.2*22.2,
we get,W=-335624.04.
now,F=W/stopping dist.
i.e F=335624.04/61
This would be -5542.95 approx
and for time you can use impulse equation,F*t=p;so t=30263.64/5543=5.45s..
for the seconf just replace the value of v;
To find the braking force, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration. In this case, the acceleration is the deceleration due to braking.
1. Find the deceleration:
We know that the car's initial speed is 50 mi/h, and it comes to a stop in a distance of 200 ft. We can convert these values to the appropriate units:
Initial speed = 50 mi/h = 50 * 5280 ft/3600 s ≈ 73.33 ft/s
Distance = 200 ft
Using the equation v^2 = u^2 + 2as, where v is final velocity, u is initial velocity, a is acceleration, and s is distance, we can solve for acceleration:
0 = (73.33 ft/s)^2 + 2a(200 ft)
a ≈ -13.61 ft/s^2 (negative because the car is decelerating)
2. Find the braking force:
Now that we have the acceleration, we can calculate the braking force using the formula F = ma, where F is force, m is mass, and a is acceleration. In this case, the force will be in the opposite direction to the motion of the car:
Mass = 3000 lb = 3000 lb * 1 slugs/32.174 ft/s^2 ≈ 93.14 slugs (1 slug ≈ 32.174 lb·s^2/ft)
Braking force = (93.14 slugs)(-13.61 ft/s^2) ≈ -1265 lb (negative because the force is in the opposite direction)
Therefore, the braking force is approximately 1265 lb.
3. Find the time required to stop:
To find the time required to stop assuming the same braking force, we can use the equation v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time:
Final velocity = 0 ft/s (since the car comes to a stop)
Initial velocity = 73.33 ft/s
Acceleration = -13.61 ft/s^2
0 = 73.33 ft/s + (-13.61 ft/s^2)t
t ≈ 5.39 s
Therefore, the time required to stop assuming the same braking force is approximately 5.39 seconds.
4. Find the distance and time required to stop if the car were initially going 25 mi/h:
Repeat steps 1 to 3 with the new initial speed of 25 mi/h to find the distance and time required to stop.