Question

A car accelerates uniformly from 18 km/h to 36 km/h in 5 sec. Calculate
(A) The acceleration
(B) The distance covered by the car at that time.

Answer 1

Hint : Since the acceleration is uniform, the uniform equation of motion can be used to calculate the distance covered. We need to concert all the units to the SI system before doing the calculation.

Formula used: In this solution we will be using the following formula;
$a = \dfrac{{v - u}}{t}$ where $a$ is the acceleration of a body, $v$ is the final velocity, $u$ is the initial velocity, and $t$ is the time taken to accelerate from the initial velocity to the final velocity.
$s = ut + \dfrac{1}{2}a{t^2}$ where $s$ is the distance covered during acceleration, $u$ is the initial velocity, and $a$ is the value of the acceleration and $t$ is the time taken to accelerate.
${v^2} = {u^2} + 2as$ where $v$ again is the final velocity, $u$ is the initial velocity, $s$ is the distance covered during acceleration from initial to final velocity, and $a$ is the acceleration.

Complete step by step answer:
To calculate the acceleration, we need to first convert the values to SI units.
$18km/h = 18\dfrac{{km}}{h} \times \dfrac{{1000m}}{{km}} \times \dfrac{1}{{3600}}\dfrac{h}{s} = 5m/s$
Also, $36km/h = 36\dfrac{{km}}{h} \times \dfrac{{1000m}}{{km}} \times \dfrac{1}{{3600}}\dfrac{h}{s} = 10m/s$
Generally, acceleration can be given by
$a = \dfrac{{v - u}}{t}$ where $a$ is the acceleration of a body, $v$ is the final velocity, $u$ is the initial velocity, and $t$ is the time taken to accelerate from the initial velocity to the final velocity.
Hence, from given values,
$a = \dfrac{{10 - 5}}{5} = \dfrac{5}{5} = 1m/{s^2}$
To calculate the distance covered in that time, we use the equation of motion
$s = ut + \dfrac{1}{2}a{t^2}$ (where $s$ is the distance covered during acceleration) which by inserting known values, we have
$s = 5\left( 5 \right) + \dfrac{1}{2}\left( 1 \right){\left( 5 \right)^2}$
$\Rightarrow s = 25 + \dfrac{{25}}{2} = 25 + 12.5$
Hence, by adding, we have
$s = 37.5m$ .

Note:
Alternatively, to calculate for the distance covered in that time, we can use the equation
${v^2} = {u^2} + 2as$ where $v$ again is the final velocity, $u$ is the initial velocity, $s$ is the distance covered during acceleration from initial to final velocity, and $a$ is the acceleration.
Hence, by inserting known values, we have
${10^2} = {5^2} + 2\left( 1 \right)s$
Hence, by making the distance subject of the formula, we have
$s = \dfrac{{{{10}^2} - {5^2}}}{2} = \dfrac{{100 - 25}}{2}$
Hence, by computing the values, we have
$s = \dfrac{{75}}{2} = 37.5m$ .