Question

A body is moving with initial velocity of 25 m/s and accelerates uniformly for 3 seconds. (4) If it attains a velocity of 34 m/s after these 3 seconds then calculate a) the acceleration and b) distance travelled during this acceleration

Answer 1

Answer:

$\boxed{\sf Acceleration = 3 \ m/s^{2}}$

$\boxed{\sf Distance \ travelled \ during \ acceleration = 88.5 \ m}$

Given:

Initial velocity (u) = 25 m/s

Final velocity (v) = 34 m/s

Time taken (t) = 3 seconds

To Find:

a) Acceleration (a)

b) Distance travelled during this acceleration (s)

Explanation:

$\sf (a) \: From \ 1^{st} \ equation \ of \ motion: \\ \sf \implies v = u + at \\ \\ \sf \implies 34 = 25 + a(3) \\ \\ \sf \implies 3a + 25 = 34 \\ \\ \sf \implies 3a = 34 - 25 \\ \\ \sf \implies 3a = 9 \\ \\ \sf \implies a = \frac{9}{3} \\ \\ \sf \implies a = 3 \: m /s^{2}$

$\sf (b) \ From \ 2^{nd} \ equation \ of \ motion: \\ \sf \implies s = ut + \frac{1}{2} a {t}^{2} \\ \\ \sf \implies s = 25 \times 3 + \frac{1}{2} \times 3 \times {(3)}^{2} \\ \\ \sf \implies s = 75 + \frac{1}{2} \times 3 \times 9 \\ \\ \sf \implies s = 75 \times \frac{1}{2} \times 27 \\ \\ \sf \implies s = 75 + 13.5 \\ \\ \sf \implies s = 88.5 \: m$