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Describe how acceleration varies kinematics problem

  • Thread starter late347
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  • #1
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Homework Statement


bicyclist increases its velocity at constant power

a) how great is this power when the bicycle's velocity increases, from 18km/h to 23km/h, during the time of 1,2 seconds

b) Within which boundaries does the acceleration vary? The combined mass of bicyclist and his bicycle is 78kg

Homework Equations


delta E= W

0,5 * m * v^2 = Ekin

The Attempt at a Solution



a)

I assumed the same mass of bycyle as it was described in the later b) portion of the problem. Assume mass = 78kg

The problem looks like very difficult to solve, without assuming the mass to be 78kg. I wonder whether it is possible to calculate the power, without this beginning assumption.

transform velocities into m/s

v0= 5 m/s (nice that it divided evenly by 3,6!)

v1= 6,388 m/s

delta Ekinetic = Work

work = 616,4552 joules
assuming flat ground during travelling.

P=W/t

P= 616,4552 joules / 1,2 s

P= 513,71 Watts

roughly 510 Watts (correct answer according to book answer, in this sense)


b) Within which boundaries does bicycles acceleration vary?

I have no idea how to strictly speaking answer the b) part

I suppose one could examine the velocities v0 and v1.

It is known that v0= 5m/s and v1= 6,388m/s

I backtracked in my textbook and found a formula which seems useful in this case. I must have glazed over this formula or simply misunderstood the formula earlier.

Ok according to my physics book one can calculate thusly

power at instant = (force at instant) * (velocity at instant)

We already calculated the power which was 510 Watts.

If we assume constant power, then we could use

P = F * v

510 W = F1* 5 m/s

510 W = F2 * 6,388 m/s

F1= 102 N
F2= 79, 8371 N

F= ma

Then we can know already the mass, and the forces F1 and F2.

We can calculate a1 and a2

F1= m*a1

F2= m * a2

a1= 102N / 78kg
= 1,3076 m/s^2

a2= 79,8371 N / 78kg
=1,0235 m/s^2
 

Answers and Replies

  • #2
416
51
You could easily derive this formula,$$ P = W / t $$
where W is work and t is time
Since (F and d are in the same direction) $$ W = F d ~~\text{if the force is constant} $$
So $$ P = \frac{F d}{t} $$
You know that ## \frac dt = v_{average}##
The formula becomes $$ P = F v_{average} $$

if you want to find the power at specific point, you could take really small intervals of d and t. However a better way is to substitute ##V_{average} ## with the ##V## at that instance.

Now for this example it seems that the force is changing, But if you take a really small interval you could approximate that the force is constant and we get this small interval by substituting the instantaneous velocity into our equation.
So assuming the question means that the power is constant at every point your solution seems to be correct at least to me.
 
Last edited:
  • #3
301
15
You could easily derive this formula,$$ P = W / t $$
where W is work and t is time
Since (F and d are in the same direction) $$ W = F d ~~\text{if the force is constant} $$
So $$ P = \frac{F d}{t} $$
You know that ## \frac dt = v_{average}##
The formula becomes $$ P = F v_{average} $$

if you want to find the power at specific point, you could take really small intervals of d and t. However a better way is to substitute ##V_{average} ## with the ##V## at that instance.

Now for this example it seems that the force is changing, But if you take a really small interval you could approximate that the force is constant and we get this small interval by substituting the instantaneous velocity into our equation.
So assuming the question means that the power is constant at every point your solution seems to be correct at least to me.
indeed the question stated that the power is constant throughout.

And the bicycle is accelerated with constant power from v0 to v1.

v0 =5m/s
v1= 6,388m/s

Would you say that, keeping in mind the pre-requisite.... constant power throughout. It seems that the force would have to be changing


so there is constant power throughout, and mass is also constant. Velocities are known, and the time to accelerate into that velocity is known.

If the power remains constant, then it means that every instance, has the same power I would say logically speaking.


Especially when we lock the power at a constant value. And we lock the instant velocities which were given for us (v0, and v1). Evidently the bicyclists mass would also not change much during the physical exertion of cycling at least in these short instances.
 

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