Question

A car accelerates uniformly from 18 km h to 36 km h in 5 min calculate acceleration and distance cover by car in that time​

Answer 1

Given ,

Initial velocity (u) = 18km/hr = 5m/s

Final velocity (v) = 36km/hr = 10m/s

Time (t) = 5min = 300secs.

Now

acceleration (a) = (v)-(u)/(t)

= 10-5/300

= 1/60 per second square

Therefore,

Distance(s) = u.t+1/2.a.t.t

= 5*300+1/2*1/60*300*300

= 2250m

Answer 2

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)/5 = 1 m/s2   Distance travelled S = u t + (1/2) a t2 = 5×5 + (1/2)×1×52 = 37.5 m

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