Question

a bus is moving with the velociity of 36 km/hr . after seeing a boy at 20 m ahead on the road, the driver applies the brake and the bus gets stopped at 10 m distance. Now, calculate acceleration as well as time taken by this bus to stop.​

Answer 1

Answer:

Assumption: the acceleration of this bus is constant while the brake was applied.

Acceleration of this bus: approximately $\left(-6.0\; \rm m \cdot s^{-2}\right)$.

It took the bus approximately $1.7\;\rm s$ to come to a stop.

Explanation:

Quantities:

• Displacement of the bus: $x = 10\; \rm m$.
• Initial velocity of the bus: $\displaystyle u = 36\; \rm km \cdot hr^{-1} = 36\; \rm km \cdot hr^{-1}\times \frac{1\; \rm m \cdot s^{-1}}{3.6\; \rm km\cdot hr^{-1}} = 10\; \rm m \cdot s^{-1}$.
• Final velocity of the bus: $v = 0\; \rm m\cdot s^{-1}$ because the bus has come to a stop.
• Acceleration, $a$: unknown, but assumed to be a constant.
• Time taken, $t$: unknown.

Consider the following SUVAT equation:

$\displaystyle x = \frac{1}{2}\, \left(a\, t^2\right) + u\, t$.

On the other hand, assume that the acceleration of this bus is indeed constant. Given the initial and final velocity, the time it took for the bus to stop would be inversely proportional to the acceleration of this bus. That is:

$\displaystyle t = \frac{v - u}{a}$.

Therefore, replace the quantity $t$ with the expression $\displaystyle \left(\frac{v - u}{a}\right)$ in that SUVAT equation:

$\displaystyle x = \frac{1}{2}\, \left(a\, \left(\frac{v -u}{a}\right)^2\right) + u\, \left(\frac{v - u}{a}\right)$.

Simplify this equation:

$\begin{aligned}x &= \frac{1}{2}\, \left(a\, {\left(\frac{v -u}{a}\right)}^2\right) + u\, \left(\frac{v - u}{a}\right) \\ &= \frac{1}{2}\left(\frac{{(v - u)}^2}{a}\right) + \frac{u\, (v - u)}{a} =\frac{1}{a}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)\end{aligned}$.

Therefore, $\displaystyle a= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)$.

In this question, the value of $x$, $u$, and $v$ are already known:

• $x = 10\; \rm m$.
• $\displaystyle u =10\; \rm m \cdot s^{-1}$.
• $v = 0\; \rm m\cdot s^{-1}$.

Substitute these quantities into this equation to find the value of $a$:

$\begin{aligned} a &= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right) \\ &= \frac{1}{10\; \rm m}\times \left(\frac{{\left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)}^2}{2} + \left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)\times 10\; \rm m \cdot s^{-1}\right)\\ &\approx -6.0\; \rm m \cdot s^{-2}\end{aligned}$.

(The value of acceleration $a$ is less than zero because the velocity of the bus was getting smaller.)

Substitute $a \approx -6.0\; \rm m \cdot s^{-2}$ (alongside $u = 10\; \rm m \cdot s^{-1}$ and $v = 0\; \rm m \cdot s^{-1}$) to estimate the time required for the bus to come to a stop:

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}}{-6.0\; \rm m \cdot s^{-2}} \approx 1.7\; \rm s\end{aligned}$.