A bus moving with a velocity 72 kmph is stopped by the application of brakes, which impact a retardation of 2.5 ms 2. (a) How long does it take for the bus to come to stop ? (b) How far does the car travel during the time brakes are applied ?.
Answers
AnswerLet t=0 be the time at which the car starts decelerating. The function x(t) will refer to the position of the car ; v(t), it’s derivative will refer to its speed ; and a(t), its second derivative will refer to its acceleration.
We choose the x axis such that x(0)=0 and v(0)=72 km/h = 72000/3600 m/s^2
Let A=2 m/s^2. We supposed that a(t) is a constant, equalling A.
We integrate a(t’) between t’=0 and t’=t, we get v(t)-v(0)=A*(t-0)=A*t, so v(t)=v(0)+A*t.
We integrate one more time to get x(t)-x(0) =v(0)*t +A*t^2/2. With x(0)=0, we have :
x(t) = v(0)*t +A*t^2/2
Now from the expression of v(t), solve for t in v(t)=0 to get the time at which the car stops moving, then inject found value of t in x(t) to get the total distance the car travelled before stopping, from the moment it started to decelerate.
Good luck !