Acceleration and distance problem

[louisG345]

Homework Statement

A car and a motorcycle at rest and 1000.0m apart start toward each other at the same time on a level track. if the car accelerates at a uniform rate of 3.7m/s^2 and the motorcycle accelerates at a uniform rate of 4.40m/s^2, at what position will they pass each other relative to the car's starting point?

Homework Equations

d=(Vi)(t)+1/2at^2
Vf^2=Vi^2+2ad
a=(Vf-Vi)/t

The Attempt at a Solution

i know that the time will be equal but i really just don't know where to go from there or what to do to solve.

 

[LowlyPion]

Homework Statement

A car and a motorcycle at rest and 1000.0m apart start toward each other at the same time on a level track. if the car accelerates at a uniform rate of 3.7m/s^2 and the motorcycle accelerates at a uniform rate of 4.40m/s^2, at what position will they pass each other relative to the car's starting point?

Homework Equations

d=(Vi)(t)+1/2at^2
Vf^2=Vi^2+2ad
a=(Vf-Vi)/t

The Attempt at a Solution

i know that the time will be equal but i really just don't know where to go from there or what to do to solve.

Homework Statement

Homework Equations

The Attempt at a Solution

Welcome to PF.

First figure how long it will take to close the distance between them to 0.

Maybe if you consider the total acceleration toward each other that can give you the time? And then with that time you can use that for the car and figure the distance it traveled when they meet?

 

[nlightner]

I would approach this problem by first identifying the given information and the unknown variable. In this case, the given information includes the initial distance between the car and motorcycle, their acceleration rates, and the fact that they start at rest. The unknown variable is the position at which they pass each other relative to the car's starting point.

Next, I would use the equations provided to determine the time it takes for the car and motorcycle to pass each other. Since they start at rest, we can use the equation a=(Vf-Vi)/t to solve for time. The car's final velocity will be the same as the motorcycle's initial velocity, since they pass each other at the same time. Therefore, we can set the car's acceleration equal to the motorcycle's initial velocity and solve for time:

3.7m/s^2 = 4.4m/s/t
t = 0.84 seconds

Now that we have the time, we can use the equation d=(Vi)(t)+1/2at^2 to solve for the distance traveled by each vehicle. The car's initial velocity is 0m/s, so we can simplify the equation to d=1/2at^2. Plugging in the values for acceleration and time, we get:

d = 1/2(3.7m/s^2)(0.84s)^2 = 1.24m

Therefore, the car and motorcycle will pass each other at a distance of 1.24m from the car's starting point. This can also be confirmed using the equation Vf^2=Vi^2+2ad, where the final velocity of both vehicles will be the same (since they pass each other at the same time) and the initial velocity for the car is 0m/s. Solving for d, we get:

d = (Vf^2-Vi^2)/2a = (4.4m/s)^2/(2*3.7m/s^2) = 1.24m

Therefore, both methods confirm that the car and motorcycle will pass each other at a distance of 1.24m from the car's starting point.