Question

A particle has initial velocity 10$m/s^2$. It moves due to constant retarding force along the line of velocity which produces retardation of $5m{s^{ - 2}}$. Then
(This question has multiple correct options.)
A) The maximum displacement in the direction of initial velocity is \[10m\].
B) The distance travelled in the first 3 seconds is \[7.5m\].
C) The distance travelled in the first 3 seconds is \[12.5m\].
D) The distance travelled in the first 3 seconds is \[17.5m\].

Answer 1

Hint: Kinematic equations are useful to solve problems in one-dimensional motion of particles with constant acceleration. These equations relate the displacement, initial velocity, final velocity, constant acceleration and time interval.

Complete step by step answer:
Given, initial velocity, \[u = 10 m/s^2\] and acceleration $a = - 5m{s^{ - 2}}$
Negative sign represents velocity is decreasing.
Maximum displacement is the direction of initial velocity means distance travelled in the initial direction until its velocity becomes zero.
Therefore, consider third equation of motion,
$ \Rightarrow {v^2} = {u^2} - 2ax$
Then $0 = {10^2} + 2( - 5)x$
$\therefore x = 10m$
Displacement is $10m$.
Displacement (not distance covered) in first 3 second is calculated using,
Second equation of motion,
$x = ut + \dfrac{1}{2}a{t^2}$
$ \Rightarrow x = 10 \times 3 + \dfrac{1}{2}( - 5){3^2}$
We get, \[x = 7.5m\].
The particle travelled in the initial direction covering $10m$ till its velocity became zero, then turned back and travelled $2.5m$. Since, net displacement in $3s$ is $ + 7.5m$.
The total distance travelled in first
\[\Rightarrow 3s = 10 + 2.5\]
\[\therefore 12.5m\]

Thus, the correct options are (A) and (C).

Additional information:
The shortest straight line distance from the initial position to final position is called displacement. Displacement is a vector quantity because it has both magnitudes as well as direction. Displacement may be zero or positive or negative.
The length of the actual path between the initial and final position is called distance. It is also called path length. It is a scalar quantity because it has the only magnitude no direction. Distance is always positive.
Retardation means negative acceleration. So, when the velocity of the body increases, then acceleration will be positive. Decreasing velocity gives us negative acceleration. If the velocity decreases with uniform time then the object is said to be under retardation.

Note:
Whenever deceleration (velocity is reducing) is given in the question use minus sign.
The magnitude of displacement may or may not be equal to the path length travelled by the object.
The negative sign in the displacement represents the direction of the displacement.