# Physics Help-Kinematic Equations

• Feb 6th 2008, 08:21 AM
charitysmama
Physics Help-Kinematic Equations
Here is the question....

A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The car accelerates at a uniform rate of 3.70 m/s^{2} and the motorcycle at a uniform rate of 4.40 m/s^{2}. (a) How much time elapses before the motorcycle overtakes the car? (b) How far will each have traveled during that time? (c) How far ahead of the car will the motorcycle be 2.00 s later? (Both vehicles are still accelerating.)

I do not even know how to begin on this one. I know for the first part I need to find the final time, but I do not know how to set it up to do that. I am stressing major about this and in desperate need of some help! Thanks in advance!
• Feb 6th 2008, 08:41 AM
wingless
Hint 1:

If an object starts to accelerate (from $\displaystyle V_0 = 0$) at a constant rate a, it'll take x distance in t time.
(The units must be equivalent. For example x=meters, t=seconds, a=m/s^2)
$\displaystyle x = \frac{1}{2}at^2$

Hint 2:

The motorcycle is 25 meters behind the car. So can we say $\displaystyle x_{car} + 25 = x_{motorcycle}$ ?
• Feb 6th 2008, 08:48 AM
charitysmama
OK.... Still not getting it unfortunately. I have literally sat and stared at this Q for hours. Seems to be flying right past me. I've done well up until now, but I just cannot seem to get this.
• Feb 6th 2008, 09:04 AM
Aryth
Let's take Wingless's hints even further:

Since the motorcycle is 25m behind the car, we can say that when the car and motorcycle meet, the motorcycle will have traveled 25m more than the car, but that makes an initial position of -25m since it was that far behind the car's position (0).

Now, we look at the kinematic equation for displacement:

$\displaystyle x_f = x_0 + v_0t + \frac{1}{2}at^2$

Now we solve it for the car and the motorcycle:

Car:

$\displaystyle x_f = 0 + 0t + \frac{1}{2}(3.7)t^2$
$\displaystyle x_f = \frac{1}{2}(3.7)t^2$

Motorcyle:

$\displaystyle x_f = -25 + 0t + \frac{1}{2}(4.4)t^2$
$\displaystyle x_f = \frac{1}{2}(4.4)t^2 - 25$

Now we see that if you meet up with someone, your final position will be exactly the same, so we now have an equivalence between the motorcycle and the car.

$\displaystyle \frac{1}{2}(3.7)t^2 = \frac{1}{2}(4.4)t^2 - 25$

Put like terms on the same side and make the 25 positive:

$\displaystyle 25 = \frac{1}{2}(4.4)t^2 - \frac{1}{2}(3.7)t^2$

Now we multiply each acceleration by $\displaystyle \frac{1}{2}$

$\displaystyle 25 = 2.2t^2 - 1.85t^2$

Combine like terms:

$\displaystyle 25 = 0.35t^2$

Divide:

$\displaystyle 71.43 = t^2$

Find t:

$\displaystyle 8.45 = t$

Therefore it takes 8.45 seconds for the motorcycle to catch up to the car.

Try b and c using these.
• Feb 6th 2008, 09:04 AM
wingless
In t seconds, the car will travel $\displaystyle x_{car} = \frac{1}{2}(3.70)t^2$ and the motorcycle will travel $\displaystyle x_{motorcycle} = \frac{1}{2}(4.4)t^2$.

Now imagine the scene. The motorcycle is 25 m behind the car and they start to accelerate. By the time the motorcycle catches up the car, the motorcycle has gone 25 meters more than the car. So $\displaystyle x_{motorcycle} = x_{car} + 25$.

Put them together,

$\displaystyle \frac{1}{2}(4.4)t^2 = \frac{1}{2}(3.70)t^2 + 25$

Solving this will give you the time $\displaystyle t$.
• Feb 6th 2008, 09:08 AM
wingless
Aryth's solution is more general. You can solve questions that include a starting velocity or displacement by using his equations :)
• Feb 6th 2008, 09:24 AM
charitysmama
Thanks so much!!

For b I got 157 m for the motorcycle and 132 m for the car. For c I got that the motorcycle will be 38 m in front of the car.... but I'm not sure that's right.
• Feb 6th 2008, 11:04 AM
Aryth
For b, you take the time at which they met, and plug it in to the final position equations for each vehicle that we derived in part a:

$\displaystyle x_{car} = \frac{1}{2}(3.7)(8.45)^2$
$\displaystyle x_{car} = 132.1m$

$\displaystyle x_{motorcycle} = \frac{1}{2}(4.4)(8.45)^2 + 25^{**}$
**(Notice we add 25 to this one as opposed to subtracting it before, when considering total distance traveled, he traveled $\displaystyle \frac{1}{2}(4.4)(8.45)^2$ starting at 0m. But we know that he started at -25m, so he traveled an extra 25m.)

$\displaystyle x_{motorcycle} = 182.1m$

For c, you use the distances calculated above (Distance for motorcycle from 0 = 157), plus the acceleration for two seconds:

$\displaystyle x_{car} = 132.1 + \frac{1}{2}(3.7)(2)^2$
$\displaystyle x_{car} = 139.5m$

$\displaystyle x_{motorcycle} = 157.1 + \frac{1}{2}(4.4)(2)^2$
$\displaystyle x_{motorcycle} = 165.9m$

$\displaystyle x_{ahead} = 165.9 - 139.5 = 26.4m$

There you are.

I hope you know when dealing with the total distance traveled you had to ADD 25, and then you had to subtract 25 to deal with the distance away from the car since:

Reference Point for Total Distance = -25
Reference Point for Distance ahead of Car = 0