Question

A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

Answer 1

GiveN :

• Initial velocity (u) = 0 m/s
• Uniform Acceleration
• Time interval (t) = 5.21 seconds
• Distance travelled (s) = 110 m

To FinD :

• Acceleration of the car

SolutioN :

As we are given that a car starts from rest and accelerates uniformly over time 5.21 seconds, It means car was initially at rest and initial velocity will be 0 m/s. For calculating acceleration, use 2nd equation of motion.

⇒s = ut + ½at²

⇒100 = 0*5.21 + ½ * a * (5.12)²

⇒100 = 0 + ½ * a * 26.21

⇒100 = (26.21 * a)/2

⇒26.21 * a = 110 * 2

⇒26.21 * a = 220

⇒a = 220/26.21

⇒a = 8.3973

⇒a ≈ 8.4

∴ Acceleration of the car is 8.4 m/s² (approx.)

_________________________

• Acceleration is Rate of change of velocity.

• It is a vector quantity.

• SI unit of acceleration is m/s².

• Acceleration = Change in velocity/Time

• Instantaneous acceleration is given by dv/dt

Answer 2

Given , A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m.

Initial velocity , u = 0 m/s

Time , t = 5.21 sec

Distance , s = 110 m

Acceleration , a = ? m/s²

Use second equation of motion , to find acceleration of the car.

⇒ s = ut + ¹/₂ at²

⇒ 110 = 0 * t + ¹/₂ * a *  (5.21)²

⇒ 110 = 13.57 a

⇒ a = 8.1 m/s²

So the acceleration of the car is 8.1 m/s²