Question

Scooter can produce maximum acceleration of 5 metre per second square its break and produces maximum retardation of 10 metre per second square find the minimum time which in which starts from rest covers a distance of 1.5 km and stops again

Answer 1

Answer:

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