Uniform accelerated motion question

In summary, the problem involves a sprinter running a race with constant acceleration and passing four posts in a straight line. The distances between each post are equal and the sprinter takes 5 seconds to go from the first post to the second and 3 seconds from the second to the third. The goal is to find the time it takes for the sprinter to run from the third to the fourth post, given that the distance between each post is 20 meters. The solution involves using the formula v^2 = u^2 + 2as to calculate the final velocity at the fourth post, and then using t = (v-u)/a to solve for the time. By calculating the average velocities for the first two time intervals and
  • #1
Nimrod 7
10
0

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)
 
Physics news on Phys.org
  • #2
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.
 
  • #3
2/3ms^2 is what I got for acceleration
Can this be solved as simultaneous equations?
 
  • #4
You seem to have done the hard part ok. Please show your working from there.
 
  • #5
haruspex said:
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
 
  • #6
Nimrod 7 said:
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.
 
  • #7
PeterO said:
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
 
  • #8
Nimrod 7 said:
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.
 
  • #9
haruspex said:
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
 
  • #10
Nimrod 7 said:
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
 
  • #11
PeterO said:
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
 
  • #12
Nimrod 7 said:
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.
 

Related to Uniform accelerated motion question

1. What is uniform accelerated motion?

Uniform accelerated motion is a type of motion in which an object moves with a constant acceleration. This means that the object's velocity changes by the same amount in each unit of time.

2. How is uniform accelerated motion different from uniform motion?

In uniform motion, an object moves with a constant velocity, meaning that its speed and direction do not change. In uniform accelerated motion, the object's velocity changes by a constant amount, resulting in a change in its speed or direction.

3. What is the equation for uniform accelerated motion?

The equation for uniform accelerated motion is v = u + at, where v represents the final velocity, u represents the initial velocity, a represents the acceleration, and t represents the time taken.

4. How can we calculate the acceleration in uniform accelerated motion?

The acceleration in uniform accelerated motion can be calculated using the equation a = (v - u)/t, where v is the final velocity, u is the initial velocity, and t is the time taken for the change in velocity to occur.

5. Can an object have a constant acceleration and still have a changing velocity?

Yes, an object can have a constant acceleration and still have a changing velocity. This is because acceleration is a measure of how much an object's velocity changes in a given time period. So, even if the acceleration is constant, the velocity can still change if the time period changes.

Similar threads

  • Introductory Physics Homework Help
Replies
23
Views
327
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
363
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top