Question

A car accelerates uniformly from \[18{\text{km}}{{\text{h}}^{{\text{ - 1}}}}\] to ${\text{36km}}{{\text{h}}^{{\text{ - 1}}}}$ in $5s$. Calculate (i) the acceleration and (ii) the distance covered.

Answer 1

Hint: To find the acceleration of the car, we have to use the first kinematic equation of motion. And for finding the distance covered we can use the third equation of motion.

Formula used: The formulae used for solving this question is given by-
$v = u + at$
${v^2} - {u^2} = 2as$
Here, $s$ is the displacement covered by a particle in $t$ time with an initial velocity of $u$, acceleration of $a$ , and $v$ is the final velocity.

Complete step by step answer:
We know from the first kinematic equation that
$v = u + at$..........(1)
According to the question, the initial velocity of the car is equal to \[18{\text{km}}{{\text{h}}^{{\text{ - 1}}}}\], the final velocity is equal to ${\text{36km}}{{\text{h}}^{{\text{ - 1}}}}$. So we have
$u = 18{\text{km}}{{\text{h}}^{{\text{ - 1}}}}$, and
$v = 36{\text{km}}{{\text{h}}^{ - 1}}$
We know that $1km = 1000m$ and $1h = 3600s$. So we can write
$u = \dfrac{{18 \times 1000}}{{3600}}m{s^{ - 1}}$
$ \Rightarrow u = 5m{s^{ - 1}}$ ……..(2)
Also
$v = \dfrac{{36 \times 1000}}{{3600}}m{s^{ - 1}}$
$ \Rightarrow v = 10m{s^{ - 1}}$ …....(3)
Also, the total time taken is equal to $5s$. So we have
\[t = 5s\]…………..(4)
Substituting (2), (3) and (4) in (1) we have
$10 = 5 + 5a$
$ \Rightarrow 5a = 5$
Dividing both sides by $5$ we ge
$a = 1m{s^{ - 2}}$….....(5)
Now, we also know that from the third kinematic equation of motion
${v^2} - {u^2} = 2as$
Substituting (2), (3), and (5) in the above equation, we have
${10^2} - {5^2} = 2 \times 1 \times s$
$ \Rightarrow 2s = 75$
Dividing both sides by $2$ we get
$s = 37.5m$
Therefore, the displacement covered by the car is equal to $37.5m$.
Since the signs of the initial and the final velocity of the car are the same, so this means that the motion of the car is unidirectional. We know that in the unidirectional motion the distance and the displacement covered are equal.
Hence, the acceleration of the car is equal to $1m{s^{ - 2}}$ and the distance covered by the car is equal to $37.5m$.

Note: For finding the distance covered by the car, we could also use the second equation of motion. Also, do not forget to convert the magnitudes of the velocities into their SI units.