Constant acceleration, bike vs. car

[Feodalherren]

Homework Statement

Vroom, vroom! As soon as a traffic light turns green, a car
speeds up from rest to 50.0 mi/h with constant acceleration
9.00 mi/h + s. In the adjoining bike lane, a cyclist
speeds up from rest to 20.0 mi/h with constant acceleration
13.0 mi/h + s. Each vehicle maintains constant velocity
after reaching its cruising speed. (a) For what time
interval is the bicycle ahead of the car? (b) By what maximum
distance does the bicycle lead the car?

Doing part a.

Homework Equations

The Attempt at a Solution

So my reasoning was that I need the position functions. I found them. I found the time at which the bike reaches its maximum velocity and selected that to be t=0. I adjusted both of the positions functions to this new setting.
Then I just went to solve for at what t they meet and my plan was just to add the (20/13) to that t and I would have the time at which the car overtakes the bike. However, I'm not successful in getting the right answer.
Where am I going wrong?

Picture of my homework posted below.

Spoiler

 

[Simon Bridge]

Did you account for velocity being in miles-per-hour and time being in seconds?
Note: You didn't need to reset the t=0 to when the bike hit top speed.

You reasoned that the bike is still ahead of the car at t=(20/13)s (did you check?)
Also that the car overtakes the bike while it is still accelerating, i.e. sometime before t=(50/9)s. (did you check?)

 

[Feodalherren]

Thanks for the quick reply.

I tried messing with them and it only made everything go wrong. I have no idea why. I know that my t=20/13 is correct because of my solutions manual.
Yes I know that the bike is ahead until then because the bike accelerates faster, so until it stops accelerating it will always be ahead.

If the t for when the car overtook the bike would be greater than 50/9 then I'd know that the car was no longer accelerating when it took over the bike and I'd have to adjust for it. But that's not the case according to the solutions manual so no need to worry about that.

As for re-adjusting for t=0 I did that because the graph of the position of the motorcycle becomes linear at that point. The only other way that I know of graphing it would be piece wise but that seems like a lot of work for no apparent reason. The car's position function remains the same because it is still accelerating, now it only has an initial velocity which I accounted for already.

 

[Simon Bridge]

Well - to find the intersection of the linear part of the bikes displacement with the quadratic part of the car's displacement, you don't need to shift the origin. What is the equation of that line? (It only has to match the bikes displacement after t=20/13s - that it is different before then does not matter).

But it is not clear that you have taken the change in units into account ... i.e. you got the equations of motion by integrating the acceleration twice ...

So you start with a=9mph/s, you go to v(t)= 9t(mph), then to x(t)=9t²/2 ... what are the units?
(eg. if this acceleration was held for 2s, did the bike travel 9(2)²/2 = 18 what?)

 

[Nickie]

Hello! First, let's start by clarifying the problem. The problem states that the acceleration for the car is 9.00 mi/h + s, which is a bit confusing. I am assuming that the correct acceleration for the car is 9.00 mi/h^2. Similarly, the acceleration for the bike should be 13.0 mi/h^2.

Now, let's move on to solving the problem. You are correct in finding the position functions for both the car and the bike. However, instead of setting t=0 at the time when the bike reaches its maximum velocity, we can set t=0 at the time when both the car and the bike start moving. This will make our calculations a bit easier.

So, the position function for the car would be x_car = 4.5t^2 and the position function for the bike would be x_bike = 6.5t^2.

(a) To find the time interval when the bike is ahead of the car, we can set the two position functions equal to each other and solve for t. This will give us the time at which they meet.

4.5t^2 = 6.5t^2
t = 0 or t = 0.692

Since t=0 is when both the car and the bike start moving, we can ignore this solution. Therefore, the bike is ahead of the car for a time interval of 0 < t < 0.692 hours.

(b) To find the maximum distance by which the bike leads the car, we can plug in t=0.692 into either of the position functions. I will use the bike's position function.

x_bike = 6.5(0.692)^2 = 3.5 miles

Therefore, the bike leads the car by a maximum distance of 3.5 miles.

I hope this helps! Keep up the good work as a scientist. Remember to always check the given information and equations to make sure they are correct before solving the problem.