Physics help - kinematics

  • Thread starter an_mui
  • Start date
  • #1
an_mui
47
0
Problem 1.

In a race a distance d long, A and B cross the finish line in a dead heat, both taking a time T. Accelerating uniformly, A takes a time ta and B a time tb to attain maximum speed, which they maintain for the rest of their race.

a) what is the acceleration of each sprinter?
b) what are their respective maximum speeds?

For both questions, express your answer in terms of d, T, ta and tb.

This is what I've done so far.. my logic is probably all wrong but any help is appreciated.

1. First, I think that the equation of A and B will be very similar (only ta, tb and maximum speeds are different).

2. I use the kinematic equations, and divide the race of A into two halves. The first half is when A is accelerating uniformly.

Phase 1
V2 = v1 + a(delta d)^2
V2 = a(delta d)^2

the distance traveled in phase 1 would be (v2 + v1)/2 x ta
d1 = (v2/2)ta

Phase 2
d2 = v2(T - ta)

Can someone tell me if my logic up to this point is correct?
 

Answers and Replies

  • #2
stunner5000pt
1,455
2
you're not using the correct formula it is

[tex] v_{2}^2 = v_{1}^2 + 2ad [/tex]
good idea doing it in phases i did it the same way. So for the first leg of A you have uniform acceleration where u use the above formula
For the second leg of A you have constant velocity with time where time is
T - Ta (do you see why?) the total time - the time taken to reach is the time A will be at cosntant velocity. Now Add the distances from both phases. And isolate for aceleration. Do the same for B.
 
  • #3
an_mui
47
0
Sorry I am stuck again.

After I added the two phases togehter, I am left with the following equation

dtotal = (v2)^2 / 2a + v2 ( T - Ta)

how would i solve for a?
 
  • #4
Fermat
Homework Helper
872
1
an_mui said:
Sorry I am stuck again.

After I added the two phases togehter, I am left with the following equation

dtotal = (v2)^2 / 2a + v2 ( T - Ta)

how would i solve for a?
I assume what you have is,

[tex]d_{total} = \frac{v_2^2}{2a} + v_2 ( T - T_a)[/tex]

where a is the acceleration during the first phase and v2 is the final velocity, with v1 = 0.

you will have gotten v2 = aTa - just substitute.
 
  • #5
an_mui
47
0
Got the help I needed, thanks!
 
Last edited:
  • #6
Fermat
Homework Helper
872
1
Yes, you did it right - those answers are correct.

btw, your answer for the acceln might look neater if the t1 was taken out as a factor :smile:

[tex]
a = \frac{2d}{t_{1}(2T - t_{1})}[/tex]
 
  • #7
an_mui
47
0
thank you very much! I really appreciate your help =)
 

Suggested for: Physics help - kinematics

  • Last Post
Replies
12
Views
1K
Replies
1
Views
620
Replies
5
Views
312
Replies
8
Views
379
J
  • Last Post
Replies
2
Views
475
  • Last Post
Replies
8
Views
584
Replies
10
Views
758
Replies
13
Views
372
Replies
2
Views
496
Top