Question

A racing car accelerates uniformly through three gears changes will for $6.0s$ what is the average speeds $20\,m{s^{ - 1}}$ for $2.0s$ , $40\,m{s^{ - 1}}$ for $2.0s$ , $60\,m{s^{ - 1}}$ for $6.0s$ what is overall average speed of car?
A. $12\,m{s^{ - 1}}$
B. $40\,m{s^{ - 1}}$
C. $13.3\,m{s^{ - 1}}$
D. $48\,m{s^{ - 1}}$

Answer 1

Hint: In order to solve this question we need to understand speed and velocity. Speed is defined as the distance covered by the body in unit interval of time, it is a scalar quantity whereas average speed is defined as total distance covered in total interval of time. Velocity is defined as the displacement per unit interval of time, it is a vector quantity as it has both magnitude and direction and also we can add the two vectors by triangle law of vector addition. Instantaneous speed is defined as a derivative of distance with respect to time.

Complete step by step answer:
Let the average speed during the first ${t_1} = 2s$ be ${v_1} = 20\,m{s^{ - 1}}$.Let the distance covered be, ${d_1}$. So using formula of speed we get,
${v_1} = \dfrac{{{d_1}}}{{{t_1}}}$
$\Rightarrow {d_1} = {v_1}{t_1}$
Putting values we get,
${d_1} = (20 \times 2)m$
$\Rightarrow {d_1} = 40\,m$
Let the average speed during second ${t_2} = 2s$ be ${v_2} = 40\,m{s^{ - 1}}$.Let the distance covered be, ${d_2}$.So using formula of speed we get,
${v_2} = \dfrac{{{d_2}}}{{{t_2}}}$
$\Rightarrow {d_2} = {v_2}{t_2}$
Putting values we get,
${d_2} = (40 \times 2)m$
$\Rightarrow {d_2} = 80\,m$

Let the average speed during the first ${t_3} = 6s$ be ${v_3} = 60\,m{s^{ - 1}}$.Let the distance covered be ${d_3}$. So using formula of speed we get
${v_3} = \dfrac{{{d_3}}}{{{t_3}}}$
$\Rightarrow {d_3} = {v_3}{t_3}$
Putting values we get,
${d_3} = (60 \times 6)m$
$\Rightarrow {d_3} = 360\,m$
So the total distance covered is,
$d = {d_1} + {d_2} + {d_3}$
Putting values we get,
$d = 40 + 80 + 360$
$\Rightarrow d = 480\,m$
And the total time taken is, $t = {t_1} + {t_2} + {t_3}$
Putting values we get, $t = 2 + 2 + 6$
$t = 10s$
So the overall average speed is, $v = \dfrac{d}{t}$
$\therefore v = \dfrac{{480}}{{10}}$
so we get, $v = 48\,m{s^{ - 1}}$

So the correct option is D.

Note: It should be remembered that instantaneous velocity can also be found from the distance time graph of motion of the body. Instantaneous velocity at a particular time is defined as the tangent of the distance time graph at that time. If and only if velocity and acceleration is uniform then we can apply the equation of motion.