Question

scooter can produce the maximum acceleration of 5 metre per second square its breaks can produce the maximum retardation of 10 metres per second square the minimum time in which it can cover a distance of 1.5 km is

Answer 1

2as=v^2-u^2  [v= final velocity; u= initial velocity; a= acceleration; s= distance] While acceleration: u=0 v=v a=5m/s^2 s=s ∴2×5×s=v^2-0^2 10s=v^2 s=v^2/10   While deceleration: u=v v=0 a=10m/s^2 s=s ∴2×10×s=0^2-v^2 20s=v^2 s=v^2/20  Now, both the distances should add up to 1500m...(1.5km=1500m)  v^2/10+v^2/20=1500 (2v^2+v^2)/20=1500 3v^2=1500 v^2=1500÷3 v=√10000 v=100  Hence, The minimum time for acceleration:  v=u-at     or v-u=at  v=100 u=0 a=5m/s^2 t=t  Therefore, 100-0=5t 100÷5=t  t=20 seconds  The minimum time for retardation:  v-u=at  v=0 u=100 a=10m/s^2 t=t      Therefore,  0-100=10t t=100÷10  t=10 seconds    Now in order to find the minimum time in which the scooter can cover a distance of 1.5km, add minimum time took for acceleration and retardation...   Therefore,              Minimum time in which the scooter can cover a distance of 1.5km= 10+20                                                                                         =30 seconds    

Answer 2

Answer:

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