Question

a car accelerates uniformly from 18 kilometre per hour to 36 kilometre per hour in 2 seconds calculate the acceleration and the distance covered by the car in that time​

Answer 1

Answer:

Acceleration = 2.5 m/s²

Distance = 15 metres

Explanation:

Given :

• Initial velocity = u = 18 km/h

• Final velocity = v = 36 km/h

• Time taken = t = 2 seconds

To find :

• Acceleration of the car

• Distance travelled by the car

Acceleration = (v-u) / t

V=36 km/h = 36×5/18 = 10 m/s

u = 18 km/h = 18×5/18 = 5 m/s

Acceleration = (10-5)/2

Acceleration = 5/2

Acceleration = 2.5 m/s²

The acceleration of the car is equal to 2.5 m/s²

Using the third equation of motion :

V²-u²=2as

10²-5²=2×2.5×s

100-25=5s

75=5s

s=75/5

s=15 metres

The acceleration of the car is 2.5 m/s² and distance is 15 metres

Answer 2

Answer :

Initial speed = u = 18 km/hr = 5 m/s

Final sped = v = 36 km/hr = 10 m/s

acceleration a = (v-u)/t = (10-5)/2 = 2.5 m/s2

Distance travelled S = ut + (1/2) at²

→ 5 × 2 + (1/2) × 1 × 2²

→ s = 15 m

$\rule{150}{2}$