Question

A car starts from rest and accelerates uniformly for 10 seconds to a velocity of 8m/s .it runs at a constant velocity and is finally brought to rest in 64m with a constant retardation. The total distance covered by the car is 584m. Find acceleration, retardation and total time taken

Answer 1

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

pls mind that(^2) mean square...........
any confusion message.
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Answer 2

ans is attached..

hope this helps ❤