CAPA problem - Kinematics in 1 Dimension

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ghostanime2001
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Homework Statement



A car accelerates at 2.10 m/s2 along a straight road. It passes two marks that are 29.6 m apart at times t=4.10 s and t=4.90 s. What was the car's velocity at t=0?

I'm assuming the car has constant acceleration of 2.10 m/s2

Given
[tex]\Delta[/tex]x = 29.6 m
a = 2.10 m/s2
vi = ?
[tex]\Delta[/tex]t = ? (I was thinking it might be 0.8 s because of 4.9 - 4.1 s )

The equation I think I should use is:
[tex]x_{f} = x_{i} + v_{i}\Delta t + \frac{1}{2}a(\Delta t)^{2}[/tex]

[tex]x_{f} - x_{i}= v_{i}\Delta t + \frac{1}{2}a(\Delta t)^{2}[/tex]

[tex]\Delta x= v_{i}\Delta t + \frac{1}{2}a(\Delta t)^{2}[/tex]

solving for [tex]v_{i}[/tex] gives the expression:

[tex]\frac{\Delta x - \frac{1}{2}a\Delta t^{2}}{\Delta t} = v_{i}[/tex]

substitute all the numbers:

[tex]\frac{(29.6) - \frac{1}{2}(1.2)\(0.8)^{2}}{((0.8)}[/tex]

[tex]36.52 m/s = v_{i}[/tex]

Am I right or wrong with this answer? Also, I couldn't check to make sure if its right or wrong because I kept on thinking the initial velocity should be zero and i held on to that and kept on asnwering that on CAPA and so as a result, I used up all of my tries :(
 
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Okay so is it like this then?

[tex]v_{f}=v_{i}+a\Delta t[/tex] from ti=0 to tf=4.1

[tex]36.52=v_{i}+(2.1)(4.1)[/tex]

[tex]-v_{i}=(2.1)(4.1)-36.52[/tex]

[tex]-v_{i}=-27.91[/tex]

[tex]v_{i}=27.91[/tex] m/s

Okay now?
 
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ghostanime2001 said:
substitute all the numbers:

[tex]\frac{(29.6) - \frac{1}{2}(1.2)\(0.8)^{2}}{((0.8)}[/tex]

[tex]36.52 m/s = v_{i}[/tex]

(

The acceleration is 2.1 m/s^2. Correct your result.

ehild
 
wow.. after so long finally I figured it out :( and I lost all my marks on that one question.
 
how do I know this is the answer?
 
It traveled 29.6m in .8 seconds and then you can change that to m/s for the velocity