Question

A body moves from rest with a uniform acceleration and travels 270m in 3s. Find the velocity and the acceleration of body

Answer 1

Answer:

$\boxed{\mathfrak{ Acceleration \ (a) = 60 \ m/s^2}}$

$\boxed{\mathfrak{Final \ velocity \ (v) = 180 \ m/s}}$

Given:

Initial velocity (u) = 0 m/s

Distance travelled (s) = 270 m

Time taken (t) = 3s

To Find:

$\sf \star$ Acceleration (a)

$\sf \star$ Final velocity (v)

Explanation:

$\sf \star \ From \ 2^{nd} \ equation \ of \ motion: \\ \boxed{ \bold{s = ut + \frac{1}{2} a {t}^{2} }}$

Substituting value of s, u & t in the equation:

$\sf \implies 270 = 0(3) + \frac{1}{2} \times a \times {(3)}^{2} \\ \\ \sf \implies 270 = \frac{1}{2} \times a \times 9 \\ \\ \sf \implies \frac{9}{2} a = 270 \\ \\ \sf \implies a = 270 \times \frac{2}{9} \\ \\ \sf \implies a = 30 \times 2 \\ \\ \sf \implies a = 60 \: m/s^2$

$\therefore$

Acceleration (a) of the body = 60 m/s

$\sf \star \ From \ 3^{rd} \ equation \ of \ motion: \\ \boxed{ \bold{ {v}^{2} = {u}^{2} + 2as}}$

Substituting value of u, a & s in the equation:

$\sf \implies {v}^{2} = {(0)}^{2} + 2 \times 60 \times 270 \\ \\ \sf \implies {v}^{2} = 2 \times 60 \times 270 \\ \\ \sf \implies {v}^{2} = 32400 \\ \\ \sf \implies v = \sqrt{32400} \\ \\ \sf \implies v = \sqrt{ {180}^{2} } \\ \\ \sf \implies v = 180 \: m/s$

$\therefore$

Final velocity (v) of the body = 180 m/s

Answer 2

Answer:

u = 0 m/s

s = 270 m

t = 3 s

• s = ut + 1/2at²

270 = 0×3 + 1/2×a×3²

a = 60 m/s²

• v = u + at

v = 0 + 60×3

v = 180 m/s