Calculating Distance and Speed in an Acceleration Chase Problem

  • Thread starter Thread starter patton_223
  • Start date Start date
  • Tags Tags
    Acceleration
AI Thread Summary
The discussion centers on a physics problem involving David driving at a constant speed of 30 m/s and Linda accelerating from rest at 2 m/s². The initial approach to finding the time it takes for Linda to catch up to David is incorrect, as it doesn't account for their respective distances traveled. To solve the problem correctly, users are advised to set up equations for the positions of both David and Linda as functions of time and equate them to find the time at which Linda passes David. The key takeaway is to use the correct equations for distance traveled under constant velocity and constant acceleration to find the point of intersection. This method ensures accurate calculations of both the distance Linda travels before passing David and her speed at that moment.
patton_223
Messages
9
Reaction score
0

Homework Statement


theres one thing i don't get about this problem

"David is driving at a steady 30/s when he passes Linda, who is sitting in her car at rest. Linda then begins to accelerate at a steady 2m/s^2 at the instant David passes her"

How far does tina drive before passing him?
What is her speed as she passes him?


Homework Equations


a=v/t
Vf^2=vi^2 +2ad
D= Vit +1/2at^2

The Attempt at a Solution



ok so i used a=v/t to solve for the amount of time it takes to catch up to david,i used 2=30/t where t would=15 seconds

I then used that 15 seconds to calculate the distance Linda has traveled using the d=vit +1/2at^2 i got 225 from that

and then i calculated her speed by using Vf^2=Vi^2 + 2ad which equaled 30m/s

I'm just not sure about that first step when i found the time, is that part right? I'm pretty sure my last 2 steps were fine, but is that 2m/s^2=30/t right?, any help would be appreciated
 
Physics news on Phys.org
patton_223 said:

Homework Statement


theres one thing i don't get about this problem

"David is driving at a steady 30/s when he passes Linda, who is sitting in her car at rest. Linda then begins to accelerate at a steady 2m/s^2 at the instant David passes her"

How far does tina drive before passing him?
What is her speed as she passes him?


Homework Equations


a=v/t
Vf^2=vi^2 +2ad
D= Vit +1/2at^2

The Attempt at a Solution



ok so i used a=v/t to solve for the amount of time it takes to catch up to david,i used 2=30/t where t would=15 seconds

I then used that 15 seconds to calculate the distance Linda has traveled using the d=vit +1/2at^2 i got 225 from that

and then i calculated her speed by using Vf^2=Vi^2 + 2ad which equaled 30m/s

I'm just not sure about that first step when i found the time, is that part right? I'm pretty sure my last 2 steps were fine, but is that 2m/s^2=30/t right?, any help would be appreciated

No! Since the time David passed Linda, David has been traveling at 30m/s. Linda has always been traveling slower than David, until Linda hits 30m/s. How can that be the moment she passes him? She must be behind him at that point. Compare the distance traveled at a steady 30m/s with the distance traveled starting from rest at an acceleration of 2m/s^2 when they are equal is when they pass. And who is Tina?
 
i see your point, but how are you supposed to get the distance if we only have 1 variable for david? do i have to calculate it in conjunction with Linda's acceleration? but I'm just confused on which equation you would use for that.
 
patton_223 said:
i see your point, but how are you supposed to get the distance if we only have 1 variable for david? do i have to calculate it in conjunction with Linda's acceleration? but I'm just confused on which equation you would use for that.

When Linda passes David, their positions are the same. So, write an expression for the position vs. time of David, write another expression for the position vs. time of Linda, and equate them. Solve for t to get the time at which Linda's position is equal to that of David's.

It's probably easiest if you take t = 0 to be the moment at which David passes Linda and she begins accelerating from rest. It's probably also easiest to call that position 0 and measure all distances from it.
 
patton_223 said:
i see your point, but how are you supposed to get the distance if we only have 1 variable for david? do i have to calculate it in conjunction with Linda's acceleration? but I'm just confused on which equation you would use for that.

Try your third equation. Linda starts from rest and accelerates at 2m/s^2. How far does she go in time t? David travels at a constant velocity of 30m/s. How far does he go in time t? For what value of t are the two distances equal? That's when they pass each other, right?
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top