Constant acceleration of an antelope

AI Thread Summary
An antelope covers 80.0m in 6.70s with a final speed of 14.6 m/s, raising questions about its initial speed and acceleration. The discussion clarifies that constant acceleration does not imply an initial speed of zero, and the correct approach involves using the formulas for displacement and velocity under constant acceleration. The initial speed can be calculated using the average speed formula, while acceleration is determined by the change in velocity over time. Participants express confusion about the concepts and seek further clarification on the equations involved. Understanding these principles is crucial for solving the problem accurately.
tAzneem
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An antelope moving with constant acceleration covers the distance 80.0m between
two points in time 6.70s. Its speed as it passes the second point is 14.6 m/s.
1) What is its speed at the first point?
2) What is the acceleration?

First I understand that when we have (constant acceleration) that means --> Vo=0m/s

*_______________* If that was the two points , so t1=0 , t2=6.70 and the all distance is 80
Vx=14.6
For 1 ) I though the answer will be zero OR can we do 80/6.70=11.9 , I really have an issue to know if the speed they are talking about is v=Δx/t or they are talking about S=D/t
2) α= Δv/Δt --> 14.6/6.70 = 2.2m/s^2
 
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tAzneem said:

An antelope moving with constant acceleration covers the distance 80.0m between
two points in time 6.70s. Its speed as it passes the second point is 14.6 m/s.
1) What is its speed at the first point?
2) What is the acceleration?

First I understand that when we have (constant acceleration) that means --> Vo=0m/s

*_______________* If that was the two points , so t1=0 , t2=6.70 and the all distance is 80
Vx=14.6
For 1 ) I though the answer will be zero OR can we do 80/6.70=11.9 , I really have an issue to know if the speed they are talking about is v=Δx/t or they are talking about S=D/t
2) α= Δv/Δt --> 14.6/6.70 = 2.2m/s^2

Homework Statement


Homework Equations


The Attempt at a Solution


Constant acceleration doesn't mean v0=0m/s. It just means the acceleration 'a' is constant. What are formulas for displacement and velocity assuming initial velocity v0 and constant acceleration a? Something like x=v0*t is only true if there is no acceleration.
 
Last edited:
Sorry , but I didn't get it , ok constant a , doesn't mean Vo=0
Can you explain the Q. more for me
 
tAzneem said:
Sorry , but I didn't get it , ok constant a , doesn't mean Vo=0
Can you explain the Q. more for me

You can fire a toy rocket forward with constant acceleration, either starting from rest or starting from a moving car. The acceleration is the same, but the two situations give different velocities; they differ by the velocity the car had at the launch point. It's not rocket science!
 
tAzneem said:
Sorry , but I didn't get it , ok constant a , doesn't mean Vo=0
Can you explain the Q. more for me

If there is no acceleration you can write the displacement as x(t)=x0+v0*t, where x0 is the initial position and v0 is the initial velocity. There is a similar expression for the case of constant acceleration. You must have seen it before. Try to find it.
 
Ok thank you both
 
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