As the maximum acceleration rate is 4 ms^-2 and the maximum velocity is 24 m/s, the time required to stop is:
tstop=adecvmax=424=6s The corresponding deceleration distance is:
Sdec=2adecvmax2=2∗4242=72m Now, we know that the car was accelerating for 56-6=50 seconds, and for the distance of 1032-72=960 meters.
This gives us:
Laccel=2aacceltaccel2
aaccel=taccel2Laccel=5022∗960=0.552m/s2 So, the maximum acceleration of the car is 0.552 m/s^2
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