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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?

kedar joshi , 13 Years ago
Grade 10
anser 1 Answers
deeksha sharma

Last Activity: 13 Years ago

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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