Question

A car starts from rest and accelerates uniformly for 10 seconds to a velocity of 5 metre per second it runs to a constant velocity and it finally brought to rest in 64 metre with the constant retardation. The total distance covered by the car is 584m. Find the value of acceleration retardation and time taken.​

Answer 1

Answer:

The car starts from rest and accelerates uniformely for 10s to a velocity of 8 m/s.The acceleration is found to be,

a=(v-u)/t=(8-0)/10=0.8m/s^2

Distance travelled during this time is,

S=0+1/2at^2

=>S=0.5x0.8x10^2

=>40m

Suppose it travels 'x' distance with constant velocity,8 m/s,for time 't'.

It then travels 64m with uniform retardation and comes to rest.

Total distance travelled=40+x+64=584

=>x=480 m

So,8 * t =480

=>t=60s

Let the car travel 64 m with uniform retardation for time t'

Using,

v^2=u^2-2as

=>0=8^2-2 (a)(64)

=>a =0.5m/s^2

retardation =0.5 m/s^2

t'=(8)/0.5=16s

Total time taken is=10s + 60s+16s=86s

Step-by-step explanation:

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