# Math Help - Kinematics in 1d question- very hard

1. ## Kinematics in 1d question- very hard

Here is the question:

A cyclist starts from rest and accelerates at $1.6ms^2$ for 10 seconds along a straight horizontal road and then continues at a constant speed. A motorist sets off from the same point as the cyclist at the same time and accelerates at a constant rate for 8 seconds, reaching a maximum speed of $V ms^1$. The car then decelerates at a constant rate until it comes to rest after a further 15 seconds. At the instant when the car comes to rest, it is overtaken by the cycle.

A) Calculate the greatest speed attained by the cyclist

B) Sketch a speed-time graph showing the motions of both the cyclist and the motorist

C) Calculate the total distanceb travelled by the cyclist before overtaking the car.

D) Calculate the value of V

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That is the question.

Now, I can do parts A and C.

A- $16ms^1$
C- 288m

I can also do half of part B, as I dont know the value of V I cannot draw the car's motion accurately, however I know the general shape!

I am having a problem with part D- calculating the value of V. I think that it is $25ms^1$; however I am unsure

Would anyone be able to clarify with me how this question is done?

cheers

2. Originally Posted by fishkeeper
Here is the question:

A cyclist starts from rest and accelerates at $1.6m/s^2$ for 10 seconds along a straight horizontal road and then continues at a constant speed. A motorist sets off from the same point as the cyclist at the same time and accelerates at a constant rate for 8 seconds, reaching a maximum speed of $V ms^1$. The car then decelerates at a constant rate until it comes to rest after a further 15 seconds. At the instant when the car comes to rest, it is overtaken by the cycle.

A) Calculate the greatest speed attained by the cyclist

B) Sketch a speed-time graph showing the motions of both the cyclist and the motorist

C) Calculate the total distanceb travelled by the cyclist before overtaking the car.

D) Calculate the value of V
this entire problem can be solved with a velocity/time graph of the cycle and the car.

the cycle's graph is a trapezoid with lower base = 23 seconds, upper base = 13 seconds, and height v = 16 m/s

the area of the trapezoid represents the cycle's displacement for the 23 second period.

$\Delta x = \frac{16}{2}(23+13) = 288$ m

the car's velocity graph is a triangle, whose area equals that of the cycle's trapezoid graph. the base of the triangle is 23 seconds and the height is v max.

$\Delta x = \frac{1}{2} v_{max} \cdot 23 = 288$

$v_{max} = \frac{2 \cdot 288}{23} \approx 25$ m/s (25.0347826...)

3. ## Can anyone advise me on sequences, nth terms etc?

Thankyou

I forgot I posted this until I saw my email

I solved this in a different way, and split the car's graph into 2 parts, for acceleration and deceleration.

I then found out what each graph held in terms of V e.g. 1/2 * V * 8 etc

then got 11.5v and divided 288 by 11.5 to get 25.04