How Long Does it Take for a Motorcycle to Overtake a Car?

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    Kinematics Racing
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Homework Help Overview

The discussion revolves around a kinematics problem involving a motorcycle overtaking a car. Participants are analyzing the time it takes for the motorcycle to overtake the car, with specific reference to the motorcycle's acceleration and maximum speed.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the motorcycle's position and the implications of its changing speed at t=7.0s. There is a focus on understanding the correct interpretation of time intervals and the contributions of different motion phases.

Discussion Status

The discussion is active, with participants providing feedback on each other's reasoning. Some guidance has been offered regarding the interpretation of time intervals and the relationship between the motorcycle's maximum speed and the overall time to overtake the car. Multiple interpretations of the problem are being explored.

Contextual Notes

There is mention of a specific answer from a book, which may influence participants' expectations. The problem context includes the transition from acceleration to constant velocity for the motorcycle, which is a key point of discussion.

rtareen
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Homework Statement
The motorcycle first takes the lead because its (constant) acceleration am = 8.40 m/s2 is greater than the car’s (constant) acceleration ac = 5.60 m/s2, but it soon loses to the car because it reaches its greatest speed vm = 58.8 m/s before the car reaches its greatest speed vc = 106 m/s. How long does the car take to reach the motorcycle?
Relevant Equations
ac = 5.6m/s2
am = 8.40m/s2
vcmax = 106m/s
vmmax = 58.8 m/s
2E0E3B09-8DA2-4F5E-9AF3-E51F178F3ED7.jpeg


Here is my attempt at the solution but I got the wrong answer. The right answer is t=16.6s. I know from the book (this is an example problem) that the motorcycle reaches its max speed at t=7.0s. But I don’t know where I made the mistake that is causing me to get the wrong answer afterwords.
 
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Your error lies in how you calculate the motorcycle position.
1576983971342.png

You're covering the time between 0 s and 7.0 seconds twice, since your xmo already covered that once.
 
gneill said:
Your error lies in how you calculate the motorcycle position.
View attachment 254545
You're covering the time between 0 s and 7.0 seconds twice, since your xmo already covered that once.

I don’t see why that’s a mistake. At t=7.0s the equation for the motorcycles position changes. It no longer has an acceleration, just a constant velocity. xmo is the starting point for this new position function.
 
rtareen said:
xmo is the starting point for this new position function.
The way it is written means that the ##v \times t## term assumes a starting time of ##t = 0##. That is not the case if it's contribution should only "kick in" at ##t = 7 sec##.
 
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rtareen said:
I don’t see why that’s a mistake. At t=7.0s the equation for the motorcycles position changes. It no longer has an acceleration, just a constant velocity. xmo is the starting point for this new position function.

It seems that your answer of ##9.6s## is the time starting from when the motorcycle reaches its maximum speed (after ##7s##). The answer of ##16.6s## is the time from when both vehicles start. I.e. your answer plus the ##7s##.

PS after ##9.6s## the speed of the car is ##53.76 m/s##, so it's not yet reached the speed of the motorcycle.
 
Last edited:
PeroK said:
It seems that your answer of ##9.6s## is the time starting from when the motorcycle reaches its maximum speed (after ##7s##). The answer of ##16.6s## is the time from when both vehicles start. I.e. your answer plus the ##7s##.

PS after ##9.6s## the speed of the car is ##53.76 m/s##, so it's not yet reached the speed of the motorcycle.

So I just need to add 7s to the answer I got and I’ll have the correct answer. I can see why now. Thanks!
 

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