# Car pass train

A train is moving parallel and adjaceent to a highway with a constant speed of 33 m/s, initially a car is 37 m behind the train, traveling in the same direction as the train at 47 m/s and accelerating at 5 m/s^2.
What is the speed of the car just as it passes the train? Answer in units of m/s
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Basically I found that:
Vo= 47 m/s
a= 5 m/s^2
x= 37 m
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Somehow i found out that
t= .869919 s
Vf= 50.7838
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However my answer is incorrect. I know the train is still moving but I cant seem to figure out how to connect the train factor and the car factor. How would I go about to solving this question?
thanks

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Hootenanny
Staff Emeritus
Gold Member
How about assuming the train is stationary and taking the initial speed of the car to be 47 - 33 = 14 m.s-1?

berkeman
Mentor
Somehow you found out that...?

Write the two equations of motion for the car and train, and apply the initial conditions that you are given. Then set the position of the car and train equal to each other, and solve for t. Then use the t to tell you the speed of the car at that moment.

I tried the problem again, this is what I did

Car:
Vo=14m/s
a= 5 mi/s^2
x=37m

Train:
Vo=33 m/s
a= 0 m/s
x=37m

Vf^2=Vo^2 +2ax
Vf^2= (14 m/s)^2 + 2((5m/s^2)(37))
Vf^2= 196m^2/s^2 +740 m^2/s^2
square root of Vf^2= square root (936 m^2/s^2)
Vf= 30.5941 m/s
Then I took this number added it to 47 the initial cars velocity and i got 77.5941.
I entered the answer and I was still incorret.

berkeman
Mentor
The general equations of motion to use are:

$$x = x_0 + v_0 t + \frac{a t^2}{2}$$

$$v = v_0 + a t$$

I'm not familiar with the $$v^2 = {v_0}^2 + 2 a x$$ equation, although that may be a valid shortcut. I would start by writing the above 2 equations for each vehicle, applying the initial conditions, and then solving the equations for equal displacements x at some time in the future. I'm pretty sure that approach will get you to the answer.