Question

A bus starts from rest with constant acceleration of $5{\text{ms}}^{-1}$. At the same time a car travelling with a constant velocity of $50{\text{ms}}^{-1}$ overtakes and passes the bus. Find at what distance will the bus overtake the car ?

• A. $200\text{m}$
• B. $500\text{m}$
• C. $1000\text{m}$
• D. $1200\text{m}$

Answer 1

The correct option is C $1000\text{m}$

Suppose the bus overtakes the car after covering distance $s$. When the two meet, time taken $t$ is same.

For bus, $u=0;a=5{\text{m/s}}^2$

Using second equation of motion,

$s=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}\times 5t^2$

$\Rightarrow s=\frac{5}{2}{t}^2.....\left(1\right)$

For car, $u=50\text{m/s};a=0$

$s=ut+\frac{1}{2}a{t}^2=50\text{t}+0=50t$

Putting the value $s$ from eq. $\left(1\right)$

$\therefore \frac{5}{2}{t}^2=50t$

$\Rightarrow t=20\text{s}$

Hence,
$s=50\text{t}=50\times 20=1000\text{m}$

So, option $\left(c\right)$ is correct.