Question

A car is at rest at a stop sign and there is another stop sign 100 m down the road. The car can brake with an acceleration of 16 m/s/s. The car accelerates with a constant rate for some time before accelerating with that braking rate to come to a stop at the second stop sign. The entire trip takes 10 seconds. Let right be positive. Don’t round until the end. Report using 3 digits. Don't enter units.

(1) With what acceleration does the car begin its motion? (in m/s/s)

(2) How much time does it accelerate with that acceleration? (in s)

(3) What is the velocity of the car when the braking begins? (in m/s)

Answer 1

here,

for acceleration period ,

let the acceleration be a1 and time be t1

the maximum veloicty , v1 = 0 + a1 * t1

v1 = a1 * t1

for deacceleration period

the initial veloicty is v1

acceleration , a2 = - 16 m/s^2

time taken , t2 = 10 - t1

using second equation of motion

v2 = v1 + a2 * t2

0 = a1 * t1 - 16 * (10 - t1) .....(1)

the total distance traveled , s = 100 m = distance traveled while accelerating + the distance traveled while deaccelerating

100 = (0 + 0.5 * a1 * t1^2) + (v1 * t2 + 0.5 * a2 * t2^2)

100 = ( 0.5 * a1 * t1^2) + ( a1 * t1 * (10 - t1) - 0.5 * 16 * (10 - t1)^2) .....(2)

from (1) and (2)

t1 = 8.75 s

a1 = 2.29 m/s^2

a)

the acceleration of car is 2.29 m/s^2

b)

the time taken while accelerating is 8.75 s

c)

the velocity of the car when the braking begins , v1 = v0 + a1 * t1

v1 = 0 + 2.29 * 8.75 m/s

v1 = 20.04 m/s