Question

A car accelerates uniformly from 36 km h to 72 km h in 5 sec calculate acceleration and distance cover by car in that time

Answer 1

This is answer. Hope it helps. Thankyou

Answer 2

Given :

Initial velocity, u = 36 km/s = 10 m/s

Final velocity, v = 72 km/h = 20 m/s

Time, t = 6 seconds

To find :

Acceleration, a &

Distance covered, s

According to the question,

➞ v = u + at

Where,

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

➞ 20 = 10 + a × 6

➞ 20 - 10 = 6a

➞ 10 = 6a

➞ 10 ÷ 6 = a

➞ 1.67 = a

So,the acceleration is 1.67 m/s².

Now,

➞ s = ut + ½ at²

Where,

s = Distance

a = Acceleration

u = Initial velocity

t = Time

➞ s = 10 × 6 + ½ × 1.67 × 6 × 6

➞ s = 60 + 1.67 × 3 × 6

➞ s = 60 + 30.06

➞ s = 90.06

So,the distance covered by the car is 90.06 meter.