Finding speed, acceleration and distance

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SUMMARY

The discussion focuses on solving a kinematics problem involving a car with constant acceleration covering a distance of 60 meters in 6 seconds, reaching a final velocity of 15 m/s. The initial speed at the first point is determined to be 10 m/s, while the constant acceleration is calculated as 2.5 m/s². Additionally, it is established that the car was 45 meters behind the first point when at rest. The conversation highlights the importance of using appropriate kinematic equations and understanding the variables involved.

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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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