# Answer on Mechanics | Relativity Question for david roy

Question #5739

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.

Expert's answer

Let's make the following denominations:

Vl = 15 m/s;

Vc - car's velosity;

Ac = 2.5 m/s²;

T1 = 3s;

We can wright down an equation:

S1 + S2 = S3, where

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

V = Ac*T2 = 2.5*14.4853 = 36.2132 m/s.

Vl = 15 m/s;

Vc - car's velosity;

Ac = 2.5 m/s²;

T1 = 3s;

We can wright down an equation:

S1 + S2 = S3, where

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

V = Ac*T2 = 2.5*14.4853 = 36.2132 m/s.

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