Uniform accelerated motion question

[Nimrod 7]

Homework Statement

A sprinter runs a race with constant acceleration throughout. During the race he passes four posts A, B, C, D in a straight line such that |AB| = |BC| = |CD| = 20m. If the sprinter takes 5 seconds to go from A to B and 3 seconds to go from B to C, find out how long, the the nearest tenth of a second, it takes him to run from C to D

Homework Equations

s = (v + u)/2 x t

The Attempt at a Solution

I have a way of doing it but it takes up almost 1 A4 page :( I did it this way and the answer was wrong, wasn't surprised cos there was loads of room for error

I did s = (v + u)/2 x t
substituted (v - at) instead of u (v = u + at, u = v-at)
And got v1 in terms of a
Then did s = (v + u)/2 x t
Substitued (u+at) instead of v
Got u2 in terms of a

v1 = u2 so I got the acceleration
I kept doing this to get the u and v for C to D and used the acceleration to calculate the time
But it was wrong

I know there's a shorter way of doing it
Can someone help me? Final answer is 2.4s (nearest tenth of a second)

 

[haruspex]

What did you get for the acceleration? I make it 2/3, so if you did not get that please show your working that far.

 

[Nimrod 7]

2/3ms^2 is what I got for acceleration
Can this be solved as simultaneous equations?

 

[haruspex]

You seem to have done the hard part ok. Please show your working from there.

 

[Nimrod 7]

You seem to have done the hard part ok. Please show your working from there.

In the previous part, I got v |AB| = 4 + 2.5a so I solved that and got speed at B is 17/3 m/s, that's the u for |BC| so

v = u + at
v = 17/3 + 2/3 (3)
v = 17/3 + 2
v = 23/3 = speed at C

Now the last part, the time from C to D
u = 23/3 m/s a = 2/3 ms^2 s = 20m

I tried to solve this with different UVAST formulae but I keep getting stuck or the wrong answer, typically something like 37 seconds which is impossible

 

[PeterO]

In the previous part, I got v |AB| = 4 + 2.5a so I solved that and got speed at B is 17/3 m/s, that's the u for |BC| so

v = u + at
v = 17/3 + 2/3 (3)
v = 17/3 + 2
v = 23/3 = speed at C

Now the last part, the time from C to D
u = 23/3 m/s a = 2/3 ms^2 s = 20m

I tried to solve this with different UVAST formulae but I keep getting stuck or the wrong answer, typically something like 37 seconds which is impossible

A key idea I would use is that when accelerating at a constant rate of acceleration, the average velocity for a time interval is the instantaneous velocity at the middle of that time interval.

NOTE: The middle refers to the middle time-wise, not the middle position.

You can calculate the average velocity for a couple of time intervals from the date given.

 

[Nimrod 7]

A key idea I would use is that when accelerating at a constant rate of acceleration, the average velocity for a time interval is the instantaneous velocity at the middle of that time interval.

NOTE: The middle refers to the middle time-wise, not the middle position.

You can calculate the average velocity for a couple of time intervals from the date given.

But how would you use the average velocity in this question?

I solved it, thanks for your help guys
I just had to use the v^2 = u^2 + 2as formula to get the final velocity at D
Then t = v-u/a and solve for t

 

[haruspex]

Now the last part, the time from C to D
u = 23/3 m/s a = 2/3 ms^2 s = 20m

I agree with that. You must have gone wrong solving the quadratic. Check the signs.

 

[Nimrod 7]

I agree with that. You must have gone wrong solving the quadratic. Check the signs.

I didn't arrive at a quadratic for this question, I simply filled that into v^2 = u^2 + 2as. I don't even know where I went wrong earlier, must've been the different fractions

 

[PeterO]

But how would you use the average velocity in this question?

I solved it, thanks for your help guys
I just had to use the v^2 = u^2 + 2as formula to get the final velocity at D
Then t = v-u/a and solve for t

For the first 5 seconds, the average speed was 4m/s [covered 20m

So 2.5 seconds into that 5 second period, the speed was 4.0 m/s

For the next 3 seconds, the average speed was 20/3.m/s [covered the next 20m in 3 seconds]

So 1.5 seconds into that time interval speed was 20/3 m/s

so in the 4 seconds between those two times, the speed inceased by 8/3 m/s so acceleration is 2/3 m/s/s

Once you have the acceleration you can calculate just about anything you want - as you did - this is just an alternate way to calculate the acceleration.

Note - although the description took 2 minutes to type out, the mental calculation involved takes about 10 seconds to do.

Note: with this acceleration, the runner will enter that third 20m with a speed of 23/3 m/s since that last section begins 1.5 seconds after speed of 20/3 m/s was reached [middle of the second interval, or 5.5 seconds after the speed of 4 m/s was reached [middle of the first interval.

then s = ut + 0.5 at² gives

20 = 23/3 t + 1/3 t²

so

60 = 23t + t² [just getting rid of the fractions]

t = -23 ± √(529 + 240)

t = etc

 

[Nimrod 7]

For the first 5 seconds, the average speed was 4m/s [covered 20m

So 2.5 seconds into that 5 second period, the speed was 4.0 m/s

For the next 3 seconds, the average speed was 20/3.m/s [covered the next 20m in 3 seconds]

So 1.5 seconds into that time interval speed was 20/3 m/s

so in the 4 seconds between those two times, the speed inceased by 8/3 m/s so acceleration is 2/3 m/s/s

Once you have the acceleration you can calculate just about anything you want - as you did - this is just an alternate way to calculate the acceleration.

Note - although the description took 2 minutes to type out, the mental calculation involved takes about 10 seconds to do.

Note: with this acceleration, the runner will enter that third 20m with a speed of 23/3 m/s since that last section begins 1.5 seconds after speed of 20/3 m/s was reached [middle of the second interval, or 5.5 seconds after the speed of 4 m/s was reached [middle of the first interval.

then s = ut + 0.5 at² gives

20 = 23/3 t + 1/3 t²

so

60 = 23t + t² [just getting rid of the fractions]

t = -23 ± √(529 + 240)

t = etc

Thanks mate :) yeah that would've been a lot easier, that's handy enough to remember

Ohh now I get how you ended up with the quadratic. Once I had u, s and a, I just used v² = u² = 2 as and after that it was just a matter of filling in the first formula t = v - u/a (v = u + at). The quadratic might've been shorter but it's easier to go wrong with it

 

[PeterO]

Thanks mate :) yeah that would've been a lot easier, that's handy enough to remember

Ohh now I get how you ended up with the quadratic. Once I had u, s and a, I just used v² = u² = 2 as and after that it was just a matter of filling in the first formula t = v - u/a (v = u + at). The quadratic might've been shorter but it's easier to go wrong with it

I much prefer to use v² = u² + 2as then follow up with t = v - u/a (v = u + at) myself.