# At what time does the motorcycle overtake the car?

1. Jun 7, 2007

### t_n_p

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I've figured it out using a method I was previously taught, but it seems that according to the official answers, the motorcycle overtakes at 32.5seconds.

Here is what I did anyhow..

Motorcycle displacement = 100t - (850)
[The 850 came from the sum total of the triangle and the trapezium]

Car displacement = 80t - 650
[Similarly the 650 came from the sum total of the triangle and the trapezium]

Equating the two...
100t-850 = 80t-650
20t = 250
t = 12.5 seconds

2. Jun 7, 2007

### Staff: Mentor

What triangle and what trapezium. What are the limits?

Not all information is given. When the motorcycle overtakes the car depends not only on velocity and acceleration, but also position/location.

Does the car pass the motorcycle, which then accelerates? If so, then both travel the same distance in the same amount of time, assuming they are traveling parallel.

3. Jun 7, 2007

### t_n_p

The full question...

Graphs of velocity time are shown below for a car and motobike travelling along the same road. The car passes the stationary motorcycle at t=0.

the triangle and trapezium are the shapes under the graph where the velocity isn't constant.

4. Jun 7, 2007

### andrevdh

I think one should assume that the car overtakes the stationary motorcycle. At the same time it pulls away.

Since you do not know during which stage the mc catches up with the c one will start by trying the triangle first and see if the the required time is less than 10 seconds. If it is more then you add the area of the triangle in as a constant and try the trapezium with a variable time. If it requires more than 5 seconds (!!!!) you know you have to add it in as a constant and then progress with the rectangular tail area with a variable time until it catches up.

5. Jun 7, 2007

### esalihm

A more quantitative solution

I guess the actual question is stated in the title of the post which is at when does the motor catch the car? (I assume that the graph is made up of straight lines)

when you see a velocity time graph and a question of this type, you should immediately recall that the are under the graph gives you the distance travelled.

since they are at the same point at the beginnig, and the car is faster than the motor at t=0, and assuming thath they move on a straight line; they are exactly at the same point when the areas under the graphs are the same.

the area under the motor up to t=15 sec is
[(60*10)/2]+[(60+80)*5/2]+=650

the area under the car up to t=15 sec is
60*15=900

so the car is still ahead of the motor
the difference between the two is 900-650=250

the difference between the speeds of the two vehicles after t=15 is 80-60=20

because time=distance/velocity
250/20= 12,5 seconds

but there was another 15 sec that I excluded by subtracting the areas before. Therefore the answer is 12,5+15=27,5 seconds

PS: The reason I subtracted the areas up to t=15 sec was that the area(distance) of the motor did not increase at a constant rate.

PS #2: the reason I did not include any units of the distances was that the units on the graph were not consistent with each other (km/h and sec.).
But this does not make any difference because you are equating the areas. Dividing them by the same number (3,6 in this case) does not affect your results.

I am afraid that there has been something wrong with your official results, and it is not less than 10 sec. certainly

Last edited: Jun 7, 2007