Question

A bus starting from rest moves with a uniform acceleration of $0.1m{{s}^{-2}}$ for $2\min$. Then, the speed acquired and the distance travelled are:
$\begin{aligned} & A)12m,720m{{s}^{-1}} \\\ & B)72m{{s}^{-1}},120m \\\ & C)12m{{s}^{-1}},720m \\\ & D)720m{{s}^{-1}},12m \\\ \end{aligned}$

Answer 1

Uniform acceleration refers to the equal change in velocity with respect to time. Speed of the bus is determined using the first law of motion. At the same time, distance travelled by the bus is calculated using the second law of motion.
Formula used:
$1)v=u+at$
$2)s=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete answer:
We are told that a bus starting from rest moves with a uniform acceleration of $0.1m{{s}^{-2}}$ for $2\min$. From this information, we are required to determine the speed acquired by the bus and the distance travelled by the bus in $2\min$. When a body is undergoing uniform acceleration, it means that velocity of the body is changing equally with respect to time. This suggests that velocity of the body is increasing gradually.
From the first law of motion, we have
$v=u+at$
where
$v$ is the final velocity of a body
$u$ is the initial velocity of the body
$a$ is the acceleration of the body
$t$ is the time travelled by the body
Let this be equation 1.
Let us use equation 1 to determine the speed of the bus. We know that
$a=0.1m{{s}^{-2}}$, is the acceleration of the bus
$u=0m{{s}^{-1}}$, is the initial velocity of the bus at rest
$t=2\min =2\times 60s=120s$, is the time, the bus moves with uniform acceleration
Clearly, using equation 1, we have
$v=u+at=0m{{s}^{-1}}+(0.1m{{s}^{-2}}\times 120s)=12m{{s}^{-1}}$
Therefore, the final velocity of the bus is equal to $12m{{s}^{-1}}$.
Now, let us move on to calculate the distance travelled by the bus in $2\min$.
From the second law of motion, we have
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
where
$s$ is the distance travelled by a body in time $t$
$u$ is the initial velocity of the body
$a$ is the acceleration of the body
Let this be equation 2.
Using equation 2, we can determine the distance travelled by the bus in $2\min$. We know that
$u=0m{{s}^{-1}}$, is the initial velocity of the bus
$a=0.1m{{s}^{-2}}$ is the uniform acceleration of the bus
$t=2\min$, is the time, the bus moves with uniform acceleration
Clearly, using equation 2, we have
$s=ut+\dfrac{1}{2}a{{t}^{2}}=(0m{{s}^{-2}}\times 120s)+\dfrac{1}{2}\times 0.1m{{s}^{-2}}\times {{(120s)}^{2}}=\dfrac{1}{2}\times 1440m=720m$
Therefore, the distance travelled by the bus in $2\min$ is equal to $720m$.
In conclusion, the speed acquired and the distance travelled by the bus are $12m{{s}^{-1}}$ and $720m$, respectively.

Hence, the correct answer is option $C$.

Note:
Students can also determine the distance travelled by the bus in $2\min$ using the third law of motion. Using the third law of motion, we have
$2as={{v}^{2}}-{{u}^{2}}\Rightarrow 2\times 0.1\times s={{12}^{2}}-{{0}^{2}}\Rightarrow s=\dfrac{144}{0.2}=720m$
Here,
$a$ is the uniform acceleration of the bus
$s$ is the distance travelled by the bus
$u$ is the initial velocity of the bus
$v$ is the final velocity of the bus