Question

A particle starts from rest and has a constant acceleration of $4m{s^{ - 2}}$ for $4sec$ . It then retards uniformly for next $8sec$ and comes to rest. Average speed of the particle during the motion is?
A. $16m{s^{ - 1}}$
B. $8m{s^{ - 1}}$
C. $24m{s^{ - 1}}$
D. None of the above

Answer 1

To begin answering the question, we must first determine the distance travelled while moving at constant acceleration $4m{s^{ - 2}}$ for $4sec$ . Because the acceleration is constant (and unknown) in the second phase of this motion, and it comes to a stop in $8sec$ , we must compute the particle's velocity once it has finished its first acceleration. We may use the same formula to determine the distance travelled again, and then add both distances at the respective accelerations. We'll also add the total time, and then plug the numbers into the average velocity formula to get the desired result.

Complete step by step answer:
We're supposed to figure out what a particle's average speed is during a motion.
The formula for calculating average speed is as follows:
${v_{av}} = \dfrac{{distance\,traveled}}{{\Delta t}}$
Where $\Delta t$ is the time interval
We know it accelerated uniformly from rest at a rate of
${a_x} = 4m{s^{ - 2}}$ for $t = 4s$
The kinematics equation would be used.
$\Delta x = {v_{0x}}t + \dfrac{1}{2}{a_x}{t^2}$
To determine how far it moves $\Delta x$ during this acceleration
Because it began at rest, the initial velocity ${v_{0x}}$ is zero, leaving us with
$\Delta x = \dfrac{1}{2}{a_x}{t^2}$
Using values that are already known:
$\Delta {x_1} = \dfrac{1}{2}\left( {4m{s^{ - 2}}} \right){\left( {4s} \right)^2} = 32m$
The acceleration is constant (and unknown) in the second phase of this motion, and it comes to a stop in $8sec$ .
Using the equation, we must determine the particle's velocity once it has completed its first acceleration.
$\Delta {x_1} = \left( {\dfrac{{{v_{0x}} + {v_x}}}{2}} \right)t$
Where ${v_x}$ is the velocity in question, and ${v_{0x}}$ is still $0$ as before:
$32m = \left( {\dfrac{{0 + {v_x}}}{2}} \right)\left( {4s} \right)$
Now we will equate for ${v_x}$
${v_x} = 16m{s^{ - 1}}$
This value shows the particle's initial velocity as it begins to accelerate negatively. We can now apply the same formula.
$\Delta {x_2} = \left( {\dfrac{{{v_{0x}} + {v_x}}}{2}} \right)t$
To calculate the distance travelled $\Delta x$
Where,
${v_x}$ is the final velocity, is $0$ (it comes to rest)
${v_{0x}}$ is $16m{s^{ - 1}}$
$t$ is $8sec$
$\Delta {x_2} = \left( {\dfrac{{16m{s^{ - 1}} + 0}}{2}} \right)\left( {8s} \right) = 64m$
As a result, the total distance covered is
Distance travelled $= \Delta {x_1} + \Delta {x_2}$
$= 32m + 64m = 96m$
And the time $\Delta t$ is
$\Delta t = 4s + 8s = 12s$
As a result, the particle's average speed is
${v_{av}} = \dfrac{{96m}}{{12s}} = 8m{s^{ - 1}}$
Average speed of the particle during the motion is $8m{s^{ - 1}}$
Therefore, the correct option is (B) $8m{s^{ - 1}}$.

Note: It should be remembered, however, that an object's average velocity tells us nothing about what happens to it between the starting and finishing points. We can't identify whether an aeroplane passenger stops or backs up before going to the back of the plane based on average velocity, for example. To gather more information, we need to look at smaller segments of the journey across shorter time periods.