Question #3864

The last car unfastens from a moving train. The train continues to move at the same constantspeed. What is the ratio of the paths covered by the train and car up to the moment of the car’s stop? Assume that the car decelerates uniformly. Solve this problem graphically.

Expert's answer

Let V (for example, V = 30 m/s) be the speed of the train.

The acceleration of the last car after it unfastened is a (for example, a = -2 m/s^{2} because its speed decreasing).

The graph shows the paths of the train and a car until the car stopped.

Train’s path can be calculated as S_{1}(t) = Vt

and the car’s path: S_{2}(t) = Vt+(at^{2})/2.

Car is moving from the moment t0 until it stops at time t_{1}.

t_{1} can be calculated from the equation for the speed of the car:

U(t_{1})=V+a*t_{1}=0, t_{1 }= -V/a = 15 s.

The acceleration of the last car after it unfastened is a (for example, a = -2 m/s

The graph shows the paths of the train and a car until the car stopped.

Train’s path can be calculated as S

and the car’s path: S

Car is moving from the moment t0 until it stops at time t

t

U(t

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