Question

12. A body moving with a constant acceleration
travels the distances 3 m and 8 m respectively in
1 s and 2 s. Calculate : (i) the initial velocity, and
(ii) the acceleration of body.
Ans. (i) 2 m s-1, (ii) 2 m s-2​

Answer 1

$\underline{\mathfrak{Given:-}}$

$\text{A body moves with a constant acceleration travels the distances 3 m and 8 m respectively in }1\ s\ \\\text{and } 2\ s.$

$\underline{\mathfrak{To\ find:-}}$

$\text{The initial velocity, and the acceleration of the body.}$

$\underline{\mathfrak{Solution:-}}$

s₁ = 3 m & t₁ = 1 s

s₂ = 8 m & t₂ = 2 s

using the equation:

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

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

${3=u \times 1+\frac{1}{2} \times a \times 1 \times 1} \\ \\{3=u+\frac{1}{2} a} \\\\ \\{u=3-\frac{1}{2} a \ \bold{ ...(i)}$

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

$8 &=u \times 2+\frac{1}{2} a \times 2 \times 2 \\\\ 8 &=2 u+\frac{1}{2} a \times 4 \\\\ 8 &=2 u+2 a\ \bold{...(ii)}$

Putting the value of 'u' from (i) in (ii):

${8=2\left(3-\frac{1}{2} a\right)+2 a} \\\\\implies {8=6-a+2 a} \\\\\implies {8=6+a} \\\\\implies {a=8-6} \\\\ \implies \boxed{a=2\ m/s^2}$

Now, putting the value of a = 2 m/s² in (ii):

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