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Tuesday, September 16, 2008

Q) A car is traveling on a straight road. The maximum velocity that can be attained by the car is 24 m/s. The maximum acceleration and decceleration it can attain is 1 m/s^2 and 4 m/s^2 respectively. Find the minimum time taken by the car to start from rest, cover a distance of 200m and stop.


A) Let the car accelerate through distance S1, and decelerate through distance S2.

S1 + S2 = 200

[v^2 - u^2 / 2a]        +  [ V^2 - U^2 / 2A ]   = 200

v^2 / 2 + (-v^2 / 8) = 200

v^2 = 320

v = sqrt(320) m/s

Also,
v = at

t = sqrt(320)

And,
V - U = AT

T = sqrt (320) / 4


Total time = t + T = sqrt (320) (5/4)

Total time = 10 sqrt(5) seconds

= 22.36 seconds


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