Car passing two markers, kinematics question

[orangegalaxies]

Homework Statement

A car accelerates at 1.80m/s2 along a straight road. It passes two marks that are 28.1m apart at times t=3.75s and t=5.00s. What was the car's velocity at t=0?

Relevant Equations

d = V1t + 0.5at^2
V2 = V1 + at

i calculated the time difference between each marker, and using the first equation, i solved for V1.
i then used this V1 as V2, time as the first marker (3.75) and used the second equation to solve for V1.

is this method correct? can somebody please verify?

 

[haruspex]

looks right.

 

[Delta2]

Yes, your method seems correct to me too.

 

[orangegalaxies]

Yes, your method seems correct to me too.

that's a relief! would you please mind sharing your answer with me so i can verify mine?

 

[PeroK]

that's a relief! would you please mind sharing your answer with me so i can verify mine?

Using the average velocity is another way to solve this problem.

 

[orangegalaxies]

Or, you could use $d = \frac 1 2 at^2$ and get an equation for $a$ in terms of the difference in positions and difference in times that you are given.

14.605 m/s?

 

[PeroK]

14.605 m/s?

Bingo!

 

[PeroK]

14.605 m/s?

The other approach is that average speed between the markers is $22.48 m/s$ (which is $28.1m$ in $1.25s$), which (for constant acceleration) must equate to the speed at the middle time $4.375s$.

Then, you can get $v_0$ from this and the acceleration.