Kinematic Equations for Catching Runner A

In summary, Runner B will catch up to Runner A after 5.24 seconds when starting the stopwatch at the same time as Runner A, or after 8.24 seconds when starting the stopwatch when Runner A is already 15 meters ahead. The distance equations for Runner A and Runner B are DA = 5t + 15 and DB = 0.5*3t^2, respectively. When setting the two equations equal to each other, a quadratic equation is formed and solving for t gives the correct answer of 5.24 seconds.
  • #1
schang
3
0
okay, this is simple and i feel ridiculous for asking, but I'm reviewing for my final and i do not remember how to do this type of problem..

Runner A runs toward a flag pole at a constant speed of 5.0 m/s. Runner B begins to run toward the flagpole at 3.0 m/s^2 when Runner A is 15 m ahead. When will Runner B catch Runner A?

I know I'll be setting two distance equations equal to each other, but beyond that I don't know what to do.
 
Physics news on Phys.org
  • #2
Have you come across any kinematic (suvat) equations?
 
  • #3
yeah, from kinematics:

D: distance traveled (m)
A: acceleration m/(s^2)
t: time (s)
D0: initial distance

D = 0.5*A*t^2 + V*t + D0

if you start your stop watch (t=0) when runner B starts to run, then runner A is already 15 m away. therefore:

for runner A:
DA = 5*t + 15

for runner B:
DB = 0.5*3*t^2

where
DA: distance traveled by runner A
DB: distance traveleed by runner B

set DA = DB, you get a quadratic.
solve the quadratic and the right answer is 5.24 s

So it takes runner B be 5.24 s to catch up to runner runner A after he/she starts to run.

If you start your stopwatch when runner A starts to run, then your time is 5.24 + 15/5 = 8.24 seconds

Just double check my math please. hope this helps.
 
  • #4
Runner B needs to cover some extra distance so are you sure about the equation for runner B?
 
  • #5
Petkovsky said:
Runner B needs to cover some extra distance so are you sure about the equation for runner B?

I think he accounted for it with Runner A ( +15)
 
  • #6
Feldoh said:
I think he accounted for it with Runner A ( +15)

Actually yes, I am sorry, I didn't see that.
 

1. What are the Kinematic (suvat) equations?

The Kinematic (suvat) equations are a set of equations that describe the motion of an object under constant acceleration. They are commonly used in physics to solve problems involving motion, and are derived from the three fundamental equations of motion.

2. What does "suvat" stand for in the Kinematic equations?

"Suvat" is an acronym for the five variables used in the equations: displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t).

3. How do you use the Kinematic equations to solve a problem?

To use the Kinematic equations, you must first identify which variables are known and which are unknown. Then, you can select the appropriate equation and plug in the values for the known variables. Finally, solve for the unknown variable using algebraic manipulation.

4. What is the difference between average and instantaneous velocity?

Average velocity is the total displacement of an object divided by the total time taken to cover that distance. Instantaneous velocity, on the other hand, is the velocity of an object at a specific instant in time. It is calculated by taking the derivative of the displacement-time graph.

5. Can the Kinematic equations be used for objects with non-constant acceleration?

Yes, the Kinematic equations can also be used for objects with non-constant acceleration. In this case, the equations would need to be modified to account for the changing acceleration. This can be done by using calculus and taking the derivative of the acceleration-time graph to find the instantaneous acceleration at a specific time.

Similar threads

Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
4K
  • Introductory Physics Homework Help
Replies
11
Views
6K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Replies
2
Views
2K
  • Introductory Physics Homework Help
2
Replies
45
Views
9K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
296
Back
Top