A car travels a distance d = 25 m in the positive x-direction in a time of t_1 =

Question

A car travels a distance d = 25 m in the positive x-direction in a time of t_1 = 23.5 s. The car immediately brakes and comes to rest in t_2 = 5s. a.) What was the car's average velocity in the horizontal direction, in meters per second, during t_1? b.) What was the acceleration, in meters per second squared, during time interval t_1, assuming the car started from rest and moved with a constant acceleration? c.) What was the car's instantaneous velocity in the horizontal direction, in meter's per second, when it began braking? d.) Using the result from part c, what was the car's horizontal component of acceleration, in meters per second squared, during the braking period?

Explanation / Answer

d = 25m, t1 = 23.5s, t2 = 5s, v1= initial velocity = 0m/s, v2 = velocity at t1

(a)Average velocity during t1 = d/t1 = 25/23.5 = 1.06m/s

(b) For a = Acceleration during t1 ,

d = v1t1 + 0.5*a*t1^2 or 25 = 0+0.5*a*23.5^2

So,  25 = 0.5*a*23.5^2 or a = 0.09m/s^2

(c) car began braking at t1, so v2 = instantaneous velocity at t1 is

V2 = v1 + a*t1 = 0+0.09*23.5

Or, v2 = 2.1m/s

(d) during braking period, car's velocity drops to zero in 5s, so it's horizontal component of acceleration = change in velocity/time taken = 0-2.1 / 5 = -0.42m/s^2

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