Question

Your bus is leaving the stop, accelerating at a constant rate of $1m{s}^{-2}$ . When the bus leaves the stop with zero velocity you are $20m$ behind it. With what minimum constant speed you must run to catch the bus? If you run at a speed more than the minimum speed, then analyse the situation.

Answer 1

Step 1: Given data

1. The acceleration of the bus when it is leaving, $a=1m{s}^{-2}$
2. The distance between you and the bus when both are at rest, $s=20m$
3. The initial velocity of the bus, $u=0m{s}^{-1}$

Step 2: Determine the distance required to be covered to catch the bus

1. Let the distance $d$ is covered by the distance before you catch the bus,
2. So you have to cover the total distance $=s+d$
3. Suppose you catch the bus with velocity $vm{s}^{-1}$ in $tsec$ .
4. Thus we can write, $s+d=vt$ ………………(1)
5. As we know the second equation of motion is $\mathit{s}\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\mathit{t}\mathbf{}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$
6. We can write the above equation as $d=0+\frac{1}{2}\left(1\right){\left(t\right)}^2$

Step 3: Substitute the value of $\mathit{d}$ from equation $\mathbf{\left(}\mathbf{1}\mathbf{\right)}$ in $\mathit{d}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\left(1\right){\left(t\right)}^{\mathbf{2}}$

$\Rightarrow vt=20+\frac{1}{2}\times 1\times t^2$

$\Rightarrow t^2-2vt+40=0$

For $t$ to be real, the roots of the above equation must be real, i.e.,

$\sqrt{{\left(2v\right)}^2-4\times 40}=0$

$\Rightarrow 4v^2-160=0$

On further solving we get,

$\Rightarrow v=\sqrt{40}m{s}^{-1}$

Final Answer:

The minimum velocity required to catch the bus is $\sqrt{\mathbf{40}}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$ .