Acar is travelling at 25m/s when it begins to slow down and eventually stops in a 98.5m a. what is the car’s acceleration?
b. at what time after the brakes were applied was the car traveling at 12.5m/s?
c. at what time did the car travel only half the stopping distance?

(a) -3.173 m/s^2

(b) 3.94 s

(c) 2.47 s

Explanation:

initial velocity, u = 25 m/s

final velocity, v = 0

distance, s = 98.5 m

(a) Let a be the acceleration of the car

Use third equation of motion

0 = 625 + 2 x a x 98.5

a = -3.173 m/s^2

(b) v = 12.5 m/s

u = 25 m/s

a = - 3.173 m/s^2

Let the time is t.

Use first equation of motion

v = u + a t

12.5 = 25 - 3.173 t

t = 3.94 s

(c) s = 98.5 / 2 = 49.25 m

u = 25 m/s

a = - 3.173 m/s^2

Let the time be t.

Let v be the velocity at this distance.

Use third equation of motion

[tex]v^{2}=25^{2}+2\times {-3.173}\ times 49.25tex]

v = 17.17 m/s

Use first equation of motion

v = u + at

17.17 = 25 - 3.173 x t

t = 2.47 s

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