Using the second equation of motion: s = ut + at²

where: s = distance covered = 20 m
u = initial velocity = 0 m/s
a = acceleration = ? m/s²
t = time = 2s

s = ut + at²
20 = 0 × 2 + × 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 + at²

where: s = distance covered = 160 m
u = initial velocity = 20 m/s
a = acceleration = ? m/s²
t = time = 4s

s = ut + at²
160 = 20 × 4 + × a × 4× 4
160 = 80 + a × 8
160 – 80 = a × 8
80 = a × 8
a = 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