Calculating Distance and Time for a Moving Car with Constant Acceleration

[PerryKid]

Homework Statement

A car starts to run from rest with acceleration 0.5 m/s². How long does it take to travel distance 60m? [sic] (I have a Russian teacher for AP Physics)

Homework Equations

V= (X_f-X_i)/(T_f-T_i)
ΔX= V_i*T+(1/2)aT²

The Attempt at a Solution

I tried to find velocity, but that didn't really work out.
0.5 m/s²=Vf-Vi/Tf-Ti
Is there an equation or some way that I can directly solve for distance from acceleration?

 

[BruceW]

ΔX= Vi*T+(1/2)aT²
from this equation, you are close to the answer already. What does Vi stand for here? and so what is its value?

 

[PerryKid]

V_i stands for initial velocity. It has an unknown value.

 

[Chestermiller]

V_i stands for initial velocity. It has an unknown value.

What does "A car starts to run from rest" mean to you with regard to the car's initial velocity?

 

[PerryKid]

I got it in class.

This was the process:

(V_f-V_i)/(T_f-T_i)=A

Or rather... ΔV/ΔT=A

Since A= 0.5 m/s², A=(0.5 m/s)/1 s

Thus, V= 0.5 m/s

The ratio is the same and thus (0.5 m)/(1 s)=(60 m)/(x s)

60 m*s = 0.5x m*s

120 s

:P

Thank you for putting up with me and with the odd grammar.

 

[gneill]

I got it in class.

This was the process:

(V_f-V_i)/(T_f-T_i)=A

Or rather... ΔV/ΔT=A

Since A= 0.5 m/s², A=(0.5 m/s)/1 s

Thus, V= 0.5 m/s

The ratio is the same and thus (0.5 m)/(1 s)=(60 m)/(x s)

60 m*s = 0.5x m*s

120 s

:P

Thank you for putting up with me and with the odd grammar.

That can't be right. After 120 seconds with an acceleration of 0.5 m/s² the car will have traveled 3.6 kilometers...that's more than slightly larger than 60 m.

What you've sort of calculated is the time it takes for the velocity to reach 60 m/s. Unfortunately, velocity is not the same as distance

Go back to your second equation (the one for ΔX) and take another look at the posts by BruceW and Chestermiller.

 

[SteamKing]

How fast do you run when you are at rest?