Can a Man Catch a Constantly Accelerating Train?

[Jess1986]

Please help with this problem. A train moves from rest with const accel. a in a straight line. A man at distance b behind the train chases it with const. speed V. Show he can catch it if V^2 >2ab, and find when he does so.
I have found formulas for distance traveled for each and equated these but cannot understand where the Vsquared comes from.
Thanks to anyone who can help. Jess x

 

[Doc Al]

Hint: When you equate those distance formulas you'll get a quadratic equation. What must be true for that quadratic to have a physically meaningful solution?

 

[kyle8921]

I would approach this problem by first understanding the basic principles of mechanics and kinematics. The given scenario involves a train moving with a constant acceleration, and a man chasing it with a constant speed. To solve this problem, we need to use equations of motion that relate distance, time, acceleration, and velocity.

The equation that relates distance, acceleration, and time is given by d=1/2at^2, where d is the distance traveled, a is the acceleration, and t is the time. Similarly, the equation that relates distance, velocity, and time is given by d=vt, where d is the distance traveled, v is the velocity, and t is the time.

In this problem, we know that the train starts from rest, so its initial velocity (u) is 0. Therefore, the equation for distance traveled by the train is d=1/2at^2. We also know that the man is chasing the train with a constant speed V, so his distance traveled is given by d=Vt.

To show that the man can catch the train, we need to equate these two distances and solve for the time (t) when they are equal. This gives us the equation 1/2at^2=Vt.

Solving for t, we get t=2V/a. Now, we need to find the condition under which the man can catch the train. This condition is given by V^2>2ab, where b is the distance between the man and the train.

Substituting t=2V/a in the equation for distance traveled by the man, we get d=V(2V/a)=2V^2/a. This distance should be equal to the distance traveled by the train, which is given by d=1/2at^2.

Equating these two distances, we get 2V^2/a=1/2at^2. Substituting t=2V/a, we get 2V^2/a=1/2a(2V/a)^2. Simplifying this equation, we get V^2=2ab, which is the condition for the man to catch the train.

To find when the man catches the train, we substitute V^2=2ab in the equation for time, t=2V/a. This gives us t=2(2ab)/a=4b, which means