Question

A body covers 26m,28m,30m,32m in ${10^{th}},{11^{th}},{12^{th}},{13^{th}}$ second respectively. The body starts
A) From rest and moves with uniform acceleration
B) From rest and moves with uniform velocity
C) With certain initial velocity and moves with constant acceleration
D) With an initial velocity and moves with constant velocity

Answer 1

Recall that velocity is the rate of change of speed with time. A body is said to be in constant velocity if the body is moving at a constant speed in the same direction. Similarly, acceleration is the rate of change of velocity with respect to time.

Complete step by step answer:
Step I:
Let ‘u' be the initial velocity and ‘a' is the acceleration.
Given is the distance S and the time.
The formula used is
${S_n} = u + \dfrac{1}{2}(2n - 1)a$ ---(i)
When distance is $26m$ , then equation (i) becomes
$\Rightarrow 26 = u + \dfrac{1}{2}\\{ 2(26) - 1\\} a$
$\Rightarrow 26 = u + \dfrac{{19a}}{2}$ ---(ii)
Step II:
Similarly substituting values for all the given distances in equation (i),
$\Rightarrow 28 = u + \dfrac{{21a}}{2}$ ---(iii)
$\Rightarrow 30 = u + \dfrac{{23a}}{2}$ ---(iv)
$\Rightarrow 32 = u + \dfrac{{25a}}{2}$ ---(v)
Step III:
Solving equations (ii) and (iii), and evaluating values of u and a,
$\Rightarrow 28 \times 2 = 2u + 21a$
$\Rightarrow 26 \times 2 = 2u + 19a$
$\Rightarrow 56 = 2u + 21a$ ---(vi)
$\Rightarrow 52 = 2u + 19a$ ---(vii)
Subtracting both equations,
$\Rightarrow 4 = 2a$
$\Rightarrow a = \dfrac{4}{2} = 2m{s^{ - 2}}$
Step IV:
Substituting value of a in equation (vi),
$\Rightarrow 56 = 2u + 21a$
$\Rightarrow 56 = 2u + 42$
$\Rightarrow 2u = 14$
$\Rightarrow u = 7m{s^{ - 1}}$

Hence, the body starts with initial velocity and moves with constant acceleration. So, Option C is the right answer.

Note:
It is to be noted that the constant acceleration and velocity are two different terms and not to be mixed. A body is said to be moving with constant acceleration if it is moving with a constant velocity in the same direction. Also if there is no force acting on the body then the acceleration of the body will be zero. In case if the forces acting on the body cancel each other, then also acceleration becomes zero. Also, constant acceleration and zero acceleration are different. Constant acceleration means that the velocity of the body is increasing or decreasing at the same rate.