Question

a car start from rest and acquire a velocity of 54 km/h in 2seconds. find
a). the acceleration
b). distance traveled by car assume motion of car is uniform

Answer 1

Initial velocity u = 0 m/s
Final velocity v = 54 km/h = 15 m/s
Time t = 2 s
Let acceleration be 'a' and distance be 's'

The car is under uniformly accelerated motion.

So, v = u + at
So, 15 = 0 + a(2)
So, a = 15/2 m/s^2
So, a = 7.5 m/s^2

Now, s = ut + 1/2 at^2
So, s = 0 + (1/2) × (15/2) × 2^2
So, s = 15 m

Thus, acceleration of car is 7.5 m/s^2 and the distance covered is 15 m

Answer 2

A .. First 54km/hr change to m/s i.e. (54 × 1000)÷(60×60) =15 m/s
a = v/s = 15/2 =7.5 m/s
B.. S =ut + 1/2at^2
15×2+ 1/2 ×7.5 ×4 = 30+15 = 45 m