Question

An object starting from rest travels 20 m in the first 2s and 160 m in next 4s. What will be the velocity after 7s from the start?

Answer 1

Using the second equation of motion: s = ut + \(\frac12\)at² 

where: s = distance covered = 20 m 

u = initial velocity = 0 m/s 

a = acceleration = ? m/s²

t = time = 2s 

s = ut + \(\frac12\)at² 

20 = 0 × 2 + \(\frac12\)× a × 2² 

20 = 0 + 2 × a 

a = 10 m/s² 

Final velocity, v after 2s

The first equation of motion: v = u + at 

where: v = final velocity = ? 

u = initial velocity = 0 m/s 

a = acceleration = 10 m/s² 

t = time = 2s 

v = u + at 

v = 0 + 10 × 2 = 20 m/s 

Now it is given that in next 4s it covers 160 m. But now the vehicle has gained some velocity. So the final velocity of the previous case will become the initial velocity in this case.

Using the second equation of motion: s = ut + \(\frac12\)at² 

where: s = distance covered = 160 m 

u = initial velocity = 20 m/s 

a = acceleration = ? m/s² 

t = time = 4s 

s = ut + \(\frac12\)at² 

160 = 20 × 4 + \(\frac12\)× a × 4× 4 

160 = 80 + a × 8160 – 80 = a × 880 = a × 8a = 10 m/s² 

This shows that acceleration is uniform.

The first equation of motion: v = u + at 

where: v = final velocity = ? 

u = initial velocity = 0 m/s 

a = acceleration = 10 m/s² t = time = 7s 

v = u + at 

v = 0 + 10 × 7 

v = 70 m/s