Question

A body moving with constant acceleration travels the distance 3m and 8m respectively in 1s and 2s. Calculate:(1)the initial velocity, and (2)the acceleration of body.

Answer 1

Initial velocity of the body is $\boldsymbol{u}=2\ \boldsymbol{m} / \boldsymbol{s}$.

The acceleration of the body is $\boldsymbol{a}=\mathbf{2}\ \boldsymbol{m} / \boldsymbol{s}^{2}$.  

The body moves at constant acceleration over two distances in respective times. The initial velocity and acceleration of the body is calculated using the equation of motion.

Given is the distance s1 = 3 m and time t1 = 1 sec

Distance s2 = 8m and time t2 = 2 sec

From the equation of motions we have, $s=u t+\frac{1}{2} a t^{2}$

$\mathrm{s}_{1}=\mathrm{u} \mathrm{t}_{1}+\frac{1}{2}\ \mathrm{a} \mathrm{t}_{1}^{2}$

$\begin{array}{l}{3=u \times 1+\frac{1}{2} \times a \times 1 \times 1} \\ {3=u+\frac{1}{2} a} \\ {U=3-\frac{1}{2} a \ldots \ldots \text { (1) }}\end{array}$

$\mathrm{S}_{2}=\mathrm{u} \mathrm{t}_{2}+\frac{1}{2}\ \mathrm{a} \mathrm{t}_{2}^{2}$

$\begin{aligned} 8 &=u \times 2+\frac{1}{2} a \times 2 \times 2 \\ 8 &=2 u+\frac{1}{2} a \times 4 \\ 8 &=2 u+2 a & \dots \ldots \ldots(2) \end{aligned}$

Equation 1 has the u value, put it in equation 2 we have,  

$\begin{array}{l}{8=2\left(3-\frac{1}{2} a\right)+2 a} \\ {8=6-a+2 a} \\ {8=6+a} \\ {a=8-6} \\ {a=2\ \frac{m}{s^{2}}}\end{array}$

Put the value of in equation 1, we get

$\begin{array}{l}{u=3-\frac{1}{2} a} \\ {u=3-\frac{1}{2} \times 2} \\ {u=3-1} \\ {u=2\ m / s}\end{array}$

Answer 2

2 m/s is the correct answer.