Question

A car accelerates uniformly from 18 km/h to 36 km/h in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.​

Answer 1

Given:

• Initial velocity (u) = 18 km/h
• Final velocity (v) = 36 km/h
• Time interval (t) = 5 s

Unknown:

• Acceleration
• Distance covered by the car in that time.

Unit Conversion:

• Initial velocity:

=> Initial velocity (u) = 18 km/h

=> Initial velocity (u) = 18 × (5/18)

=> Initial velocity (u) = 90/18

=> Initial velocity (u) = 5 m/s

• Final velocity:

=> Final velocity (v) = 36 km/h

=> Final velocity (v) = 36 × (5/18)

=> Final velocity (v) = 180/18

=> Final velocity (v) = 10 m/s

Solution:

• Acceleration (a) :

=> Acceleration = (v - u)/t

=> Acceleration = (10 - 5)/5

=> Acceleration = 5/5

=> Acceleration = 1 m/s²

• Distance (s) :

By applying third kinematical equation of motion:

=> v² - u² = 2as

=> (10)² - (5)² = 2 × 1 × s

=> 100 - 25 = 2s

=> 75 = 2s

=> s = 75/2

=> s = 37.5 m