Finding speed, acceleration and distance

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A car with constant acceleration covers 60m in 6 seconds, reaching a velocity of 15m/s at the second point. The speed at the first point is calculated to be 10m/s, with a constant acceleration of 2.5m/s². The car was determined to be 45m behind the first point when at rest. Some participants noted that the equations used were inappropriate for the scenario, suggesting potential missing information. Ultimately, the discussion clarified that the system is not underdetermined, allowing for the correct application of kinematic equations.
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Homework Statement



A car moving with a constant acceleration covers the distance between two points 60m apart in 6.0s. Its velocity as it passes the second point is 15m/s. a) What was the speed at the first point? b) what is the constant acceleration? c) How far behind the first point was the car at rest?

Homework Equations



v=\Deltax/\Deltat
a=\Deltav/\Deltat
d=1/2at^{}2

The Attempt at a Solution


a) 60m/6s = 10m/s
speed at first point is 10m/s
b) (15m/s)/6s = 2.5m/s^{}2
constant acceleration is 2.5m/s^{}2
c) 1/2(-2.5m/s^{}2)(6.0s)^{}2 = -45m
car at rest is 45m from first point
 
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applepie89 said:

Homework Statement



A car moving with a constant acceleration covers the distance between two points 60m apart in 6.0s. Its velocity as it passes the second point is 15m/s. a) What was the speed at the first point? b) what is the constant acceleration? c) How far behind the first point was the car at rest?

Homework Equations



v=\Deltax/\Deltat
a=\Deltav/\Deltat
d=1/2at^{}2

The Attempt at a Solution


a) 60m/6s = 10m/s
speed at first point is 10m/s
b) (15m/s)/6s = 2.5m/s^{}2
constant acceleration is 2.5m/s^{}2
c) 1/2(-2.5m/s^{}2)(6.0s)^{}2 = -45m
car at rest is 45m from first point

V = d/t only if there is constant acceleration. It this case, it is not.

d=1/2at^2 only if v_initial is = 0. This is also not true.

More relevant equations are:

vf^2 = vi^2 + 2ad
d = 1/2at^2 + (vi)t

However, in both equations you still have two unknown which will yield infinite possible solutions (no specific answer). I suspect there is information missing.
 
You need to study your kinematic equations a little more and understand all the variables, you are using them wrong.
 
Krogy said:
V = d/t only if there is constant acceleration. It this case, it is not.

d=1/2at^2 only if v_initial is = 0. This is also not true.

More relevant equations are:

vf^2 = vi^2 + 2ad
d = 1/2at^2 + (vi)t

However, in both equations you still have two unknown which will yield infinite possible solutions (no specific answer). I suspect there is information missing.

? The system is not under determined. Those 2 equations right there have 2 unknowns each (vi, a) We have vf, d, and t.
 
Xerxes1986 said:
? The system is not under determined. Those 2 equations right there have 2 unknowns each (vi, a) We have vf, d, and t.

Ahh ok, I missed the Vf. In that case, we can solve for the second equation by substituting.
 
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