Question

a) A bus leaves a stop and accelerates at a constant rate for 5.0 seconds. During this time the bus travels 25 meters. After this the bus travels at a constant speed for 15 seconds. Then the driver notices a red light 18 meters ahead and applies brakes with acceleration $a_b$. Assume that the bus decelerates at a constant rate and comes to a stop sometime later just at the light.
1. What was the initial acceleration of the bus?
2. What was the velocity of the bus after 5 seconds?
3. Calculate $a_b$ .
4. How long did does the bus brake?
b) Two athletes Usha and Shiney are playing athletic games. Usha is running at a constant velocity towards Shiney who is stationary. When Usha is 12 meters away from Shiney, Shiney starts to accelerate at a constant rate of $1.5m{s}^{-2}$.
1. What is the minimum velocity with which Usha needs to run in order to just catch up with Shiney?
2. How long does Usha take to catch up with Shiney?

Answer 1

a)
1. $a=\frac{2s}{t}=\frac{2\left(2s\right)}{(5{)}^2}=2m/s^2$
$Now,a=2m/s2=>s1=25m$

2. $v=a\times t=2\times 5=10m/s=>s2=150m$
3. $a=\frac{-v^2}{2s}=\frac{-1}{2}\times \frac{{10}^2}{18}=-2.78m/s^2$ It is negative
4. $18=\frac{1}{2}\times 2.78\times t^2=>t=3.60sec$
$Also,s_3=17.98\approx 18m$
b)
$v_u=const$
$a_s=1.5m/s$
$x_u{x}_s=12m$
Usha catches up with Shiney after time t
$x_u=v_u\times t$
$x_s=0.5as.t_2$
$v_{u}t0.75t_2=12$
at time t, $v_u=v_s=1.5t$ (since Usha is over taking Shiney)
$1.5t_{2}0.75t_2-12=0$
$0.75t_2=12$
$t=4sec$
$v_u=at=6m/s$