Question

A car starts from rest and accelerates uniformly to a speed of 72km/h over 500m. if a further acceleration raises the speed of 90km/h in 10 seconds find this acceleration and the further distance moved​

Answer 1

Explanation:

Given -

Two type of acceleration :-

(i) u = 0 km/h, v = 72 km/h, distance (s) = 500 m

(ii) u = 72 km/h, v = 90 km/h, t = 10 s

To find -

Second (ii) acceleration and distance moved

Solution -

Second acceleration :-

We have to change speeds into m/s, so

72 km/h = 72×5/18 m/s

= 360/18 m/s

= 20 m/s = u

90 km/h = 90×5/18 m/s

= 450/18 m/s

= 25 m/s = v

acceleration = (v-u)/t

= (25-20)/10 m/s²

= 5/10 m/s²

= 0.5 m/s²

so acceleration (a) = 0.5 m/s².

Distance moved =

s = ut + 1/2at²

= (20×10)+ (1/2×0.5×10²)

= 200+(1/2×1/2×100)

= 200+(1/4×100)

= 200+25

= 225 m

Hence further distance = 225 m.

The acceleration = 0.5 m/s² and distance moved = 225 m.

hope it helps.