Question

a body travels 200 CM in the first 2s and 220 CM in the next 5s.Calculate the velocity at the end of the seventh second from the start​

Answer 1

Answer:

First try to find the acceleration and the initial velocity of the body. We will have to assume that the acceleration is uniform.

For the first 2 seconds,

s = distance = ut + (1/2)at^2 = 200 cm = 0.2 m

where t = time = 2 s, u = initial velocity and a = acceleration.

0.2 = 2u + 0.5a(2 ^ 2) = 2u + 2a

=> u + a = 0.1

=>u = 0.1 - a → equation 1

For the first 6 seconds (2 + 4),

s = distance = ut + (1/2)at^2 = 200 + 220 cm = 420 cm = 0.42 m

t = 6 s

0.42 = u(6) + 0.5a(6^2)

=> 0.42 = 6u + 18a

=> 0.07 = u + 3a

We’ll substitute u as 0.1 - a from equation 1

0.07 = 0.1 - a + 3a

=> 2a = 0.07 - 0.1 = -0.03

=> a = -0.03/2 = -0.015 m/s^2

u = 0.1 - (-0.015) = 0.115 m/s

Now we have the initial velocity, time and acceleration, we need to find the final velocity. So we’ll just plug in the values in this equation:

final velocity = v = u + at

v = 0.115 + (-0.015)(7)

= 0.1 m/s or 10cm/s

was that correct???