Finding speed, acceleration and distance

[applepie89]

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

 

[Krogy]

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.

 

[Xerxes1986]

You need to study your kinematic equations a little more and understand all the variables, you are using them wrong.

 

[Xerxes1986]

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.

 

[Krogy]

? 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.