# Kinematics Question -- Two motorbikes accelerating at each other...

#### michaelgtozer

1. Homework Statement
Two stationary motorbikes at time 0s are separated by distance of 80m. Both motorbikes accelerate at a constant rate as the travel towards each other. Motorbike X accelerates at a rate of 4.6m/s2 . Motorbike Y accelerates at a rate of 2.4m/s2 . Calculate the amount of time that elapses until the two motorbikes reach a distance of separation of 24m. Recall that, for an initial velocity of 0, d = 0.5at2

2. Homework Equations
d = 0.5at^2

3. The Attempt at a Solution
I know the answer to this problem is 3.5s. I am just wondering how to find a general solution for this problem, meaning a equation or a combination of equations that when the equation is solved for "t", i am able to plug in different values and get the correct answer each time.

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#### Buzz Bloom

Gold Member
Hi mike:

Your description of your solution attempt is hard to read. In particular:
I figured this out by adding 4.6 and 2.4, which says that since they are doing constant acceleration that every second they are travelling 7 meters toward each other each second, and 7 meters multiplied by 8 would give me the 56m total that they need to travel towards each other.
I suggest you express your thoughts in equations. What is the rate the two motorbikes are accelerating towards each other? How much time will it take for the bikes ti change their distance from 80 m to 24 m? Can you use the equation in (2) to represent this?

Regards,
Buzz

#### michaelgtozer

That seems to be where im having the trouble. I can't see how I can use the equation from part two get to the answer of 3.5s. When i look at the equation I see that it only allows me to put in 1 acceleration when the problem is giving me two to work with.

#### berkeman

Mentor
That seems to be where im having the trouble. I can't see how I can use the equation from part two get to the answer of 3.5s. When i look at the equation I see that it only allows me to put in 1 acceleration when the problem is giving me two to work with.
Start by defining an axis, in this case you can call it the "x-axis" with the left motorbike at the origin x=0 and the right motorbike at x=_____

Then use the full equation for the motion:

$$x(t) = x_0 + v_{0x}t + \frac{1}{2}a_x t^2$$

Write that equation for each motorbike (calling them different "x" values like "x_left(t)" and "x_right(t)", and start to solve for things...

#### michaelgtozer

Start by defining an axis, in this case you can call it the "x-axis" with the left motorbike at the origin x=0 and the right motorbike at x=_____

Then use the full equation for the motion:

$$x(t) = x_0 + v_{0x}t + \frac{1}{2}a_x t^2$$

Write that equation for each motorbike (calling them different "x" values like "x_left(t)" and "x_right(t)", and start to solve for things...
So far I have this:

I started with both bikes on the x axis:
bike x starts at 0m
bike y starts at 80m

equation for bike R is R: = 1/2at^
equation for bike L is: L = 80 - 1/2at^2

so i know both accelerations, and im trying to solve for time, but i still have 2 unknown values of x and y...
So since there are 3 unknowns it means i would need 3 equations, but i dont have a third one

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#### berkeman

Mentor
equation for bike y is: y = 80 + 1/2at^2
The right-hand motorbike accelerates to the left, which is a negative acceleration on the x-axis.

BTW, maybe call the 2 motorbikes L and R to avoid confusion with any x-axis (and any y-axis, which does not enter into this problem).

#### michaelgtozer

The right-hand motorbike accelerates to the left, which is a negative acceleration on the x-axis.

BTW, maybe call the 2 motorbikes L and R to avoid confusion with any x-axis (and any y-axis, which does not enter into this problem).
Okay so i changed the variables to R and L, also changed the sign of the acceleration in the L equation to a negative.
Any hints on what I should do next? Still very confused.

#### berkeman

Mentor
Can you show the two equations as you have them now for the motion of the two motorbikes?
Calculate the amount of time that elapses until the two motorbikes reach a distance of separation of 24m.
And what is the equation that you can write for the condition mentioned above? You would use those 3 equations to solve for the time...

#### michaelgtozer

Can you show the two equations as you have them now for the motion of the two motorbikes?

And what is the equation that you can write for the condition mentioned above? You would use those 3 equations to solve for the time...
So the two equations I have now are:
R = 1/2at^2
L = 80 - 1/2at^2

so for the third equation i think it would look something like this..
R = 80 - (24+L), so it would simplify to R = 56 - L i think..

From this would i solve R for t^2, then put in the t^2 in the L equation, then solve the 3rd equation for L, and sub it in the same L equation, then solve the entire equation for R, then put that R value in and solve for t^2?

#### berkeman

Mentor
So the two equations I have now are:
R = 1/2at^2
L = 80 - 1/2at^2

so for the third equation i think it would look something like this..
R = 80 - (24+L), so it would simplify to R = 56 - L i think..

From this would i solve R for t^2, then put in the t^2 in the L equation, then solve the 3rd equation for L, and sub it in the same L equation, then solve the entire equation for R, then put that R value in and solve for t^2?
That was a little confusing for me, so let me try to say it a bit differently.

L = left motorbike, starting at x_L(0s)=0m
R = right motorbike, starting at x_R(0s)=80m
Motorbike X accelerates at a rate of 4.6m/s2 . Motorbike Y accelerates at a rate of 2.4m/s2
So equation for the L motorbike is x_L(t) = 0 + 0 + 1/2 (4.6m/s^2)t^2
And for the R motorbike is x_R(t) = 80m + 0 - __________
And the problem asks for the time t when x_R(t) - x_L(t) = 24m

Can you use that notation and fill in the rest of the equation for x_R(t) and continue with the solution...?

#### michaelgtozer

That was a little confusing for me, so let me try to say it a bit differently.

L = left motorbike, starting at x_L(0s)=0m
R = right motorbike, starting at x_R(0s)=80m

So equation for the L motorbike is x_L(t) = 0 + 0 + 1/2 (4.6m/s^2)t^2
And for the R motorbike is x_R(t) = 80m + 0 - __________
And the problem asks for the time t when x_R(t) - x_L(t) = 24m

Can you use that notation and fill in the rest of the equation for x_R(t) and continue with the solution...?
Okay so the equation for the motion of R motorbike is R(t) = 80m + 0 - 1/2(2.4m/s^2)t^2

Now that I have all 3 equations but i do this:
x_L(t) = 0 + 0 + 1/2 (4.6m/s^2)t^2 ----> is equal to t^2 = x_L(t)/2.3

So when I combine that equation and x_R(t) - x_L(t) = 24m the final equation should be:

24 - x_L(t) = 80 - 1/2x_L(t)*(1/2.3)

and i would solve the equation for x_L(t) it looks like:
x_L(t) = -56/(-1/4.6 + 1) which i dont think is correct. The final answer to this problem should be t = 4 seconds, but i just cant seem to get that answer

#### berkeman

Mentor
So when I combine that equation and x_R(t) - x_L(t) = 24m the final equation should be:

24 - x_L(t) = 80 - 1/2x_L(t)*(1/2.3)
I'm also not quite tracking that...

To summarize:
x_L(t) = 0 + 0 + 1/2 (4.6m/s^2)t^2
x_R(t) = 80m + 0 - 1/2(2.4m/s^2)t^2
x_R(t) - x_L(t) = 24m
The final answer to this problem should be t = 4 seconds, but i just cant seem to get that answer
When I plug the first 2 equations into the 3rd equation, I get an answer which is within a percent or so of 4 seconds...What do you get? (please show each step on a separate line...)

#### michaelgtozer

I'm also not quite tracking that...

To summarize:
x_L(t) = 0 + 0 + 1/2 (4.6m/s^2)t^2
x_R(t) = 80m + 0 - 1/2(2.4m/s^2)t^2
x_R(t) - x_L(t) = 24m

When I plug the first 2 equations into the 3rd equation, I get an answer which is within a percent or so of 4 seconds...What do you get? (please show each step on a separate line...)
So when i put the first 2 equations in the 3rd i get
80 - 1.2t^2 - 2.3t^2 = 24
-3.5t^2 = -56
t^2 = -56/-3.5
t= sqrt(56/3.5)
t = 4

#### berkeman

Mentor
So when i put the first 2 equations in the 3rd i get
80 - 1.2t^2 - 2.3t^2 = 24
-3.5t^2 = -56
t^2 = -56/-3.5
t= sqrt(56/3.5)
t = 4
Yep, that's what I got.

So to recap, for these kinds of problems, set up a coordinate system, write out the full form of the Kinematic Equations that you may need to use, be careful with your algebra, and think about what is going on physically to keep the equations real. Oh, and it's a good idea to carry units along in your equations to be sure you haven't dropped a term along the way.

Good!

#### michaelgtozer

Yep, that's what I got.

So to recap, for these kinds of problems, set up a coordinate system, write out the full form of the Kinematic Equations that you may need to use, be careful with your algebra, and think about what is going on physically to keep the equations real. Oh, and it's a good idea to carry units along in your equations to be sure you haven't dropped a term along the way.

Good!
Thanks!

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