Question

A car starting from rest has a speed of $30\,kmh{r^{ - 1}}$ at any one instant. Two seconds later its speed is $36\,kmh{r^{ - 1}}$ and two seconds after that it is $42\,kmh{r^{ - 1}}$.What is its acceleration in $m{s^ - }^2$?

Answer 1

Hint: This problem can be solved using the first equation of motion. First convert the given speeds and calculate acceleration using the equation of motion. The car starts from rest, hence its initial velocity will be 0. After two seconds the velocity is given. Also, the thing to do first is to convert all the given quantities in standard SI units.

Formula Used:
To solve the problem we have used the following formula,
$v = u + at$$v = u + at$
Where $v$ represents the final velocity, $u$ represents the initial velocity, $a$ represents acceleration and $t$ represents time taken.

Complete step by step answer:
In the question it is given that when the car starts from rest its speed is $30kmh{r^{ - 1}}$ which means its initial velocity is $30kmh{r^{ - 1}}$. Two seconds later its speed is $36kmh{r^{ - 1}}$and two seconds after that it is $42kmh{r^{ - 1}}$. So, we have to first calculate the acceleration in $m{s^ - }^2$. Firstly, we will have to convert the speeds from $kmh{r^{ - 1}}$ to$m{s^{ - 1}}$. So, converting each of the given velocities we can write,

${u_1} = 30kmh{r^{ - 1}} \\
 \Rightarrow{u_1} = \dfrac{{30 \times 1000}}{{3600}}m{s^{ - 1}} \\
 \Rightarrow{u_1} = \dfrac{{300}}{{36}}m{s^{ - 1}} \\
 \Rightarrow{u_1} = \dfrac{{25}}{3}m{s^{ - 1}}$

$ \Rightarrow{u_2} = 36kmh{r^{ - 1}} \\
 \Rightarrow{u_2} = \dfrac{{36 \times 1000}}{{3600}}m{s^{ - 1}} \\
 \Rightarrow{u_2} = \dfrac{{360}}{{36}}m{s^{ - 1}} \\
 \Rightarrow{u_2} = 10m{s^{ - 1}}$

$ \Rightarrow{u_3} = 42kmh{r^{ - 1}} \\
 \Rightarrow{u_3} = \dfrac{{42 \times 1000}}{{3600}}m{s^{ - 1}} \\
 \Rightarrow{u_3} = \dfrac{{420}}{{36}}m{s^{ - 1}} \\
 \Rightarrow{u_3} = \dfrac{{35}}{3}m{s^{ - 1}}$

Now, for calculating acceleration from the formula we use the values provided in the question as,
$u = {u_1} = \dfrac{{25}}{3}m{s^{ - 1}}$
$ \Rightarrow v = {u_2} = 10m{s^{ - 1}}$
$ \Rightarrow t = 2s$........................(given in the question)
Therefore, substituting the respective values of $u,v,t$ in the formula we get,
$v = u + at$
$ \Rightarrow 10 = \dfrac{{25}}{3} + 2a$
$ \Rightarrow 2a = 10 - \dfrac{{25}}{3}$
$ \Rightarrow 2a = \dfrac{{30 - 25}}{3} \\
\Rightarrow 2a= \dfrac{5}{3}$
$ \Rightarrow a = \dfrac{5}{{3 \times 2}} \\
\therefore a= \dfrac{5}{6}m{s^ - }^2$

Hence, acceleration is $\dfrac{5}{6}m{s^ - }^2$.

Note: This problem requires conversion of units many times. So, the conversions must be done carefully so that all the quantities are in the same system of units. Also, it is important to understand which of the given velocity represents initial velocity and which one represents final velocity. While doing problems involving Newton’s equations of motion, we have to keep in mind the sign conventions. If a body is accelerating then its acceleration is positive and if the body is coming to stop after some time then it is accelerating and its acceleration is negative.