Answer to Question #5739 in Mechanics | Relativity for david roy
a lorry moving at constant 15 m/s passes a stationary car. after 3 seconds the car starts moving at an accelaration of 2.5 m/s squared in pursuit of the lorry. find the distance covered and time taken for the car to catch up with the lorry. also find the velocity with which the car overtakes the lorry.
S1 = Vl*T1 - distance that lorry had moved before car started S2 = Vl*T2 - distance that lorry had moved after car started S3 = (Ac*T2²)/2 - distance covered by car and T2 is the car moving time. So,
Vl*T1 + Vl*T2 = (Ac*T2²)/2
(Ac*T2²)/2 - Vl*T2 - Vl*T1 = 0
2.5*T2²/2 - 15*T2 - 15*3 = 0
and we've got a quadratic equation
1.25*T2² - 15*T2 - 45 = 0
Taking in account that T2>0 we obtain
T2 = (15+sqrt(15²+4*1.25*45)) / (2*1.25) ≈ 14.4853 s.
The distance covered by car is
D = (Ac*T2²)/2 = 2.5*14.4853²/2 = 262.2792 m
and the velocity with which the car overtakes the lorry is