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a car is travelling on a straight road. the maximum velocity of the car can be 24m/s. the max. acceleration and max. deceleration it can attain are 1m/(second)square and 4m/(second)square respectively. the shortest time the car takes from rest to rest in a distance of 200 mts is?
The formula is
V(max) = (abt)/(a+b)
Here,a= Max acc
b= max decc
t= total time taken
Solving, we get 24=(1*4*t)/(4+1)
24*5=4t
t=30s
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