Solving Kinematic Questions: Velocity at t=0

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The discussion revolves around determining a car's velocity at t=0, given its acceleration and the distance covered between two time markers. The initial attempt used the kinematic equation but resulted in calculating the velocity at t=3.95s instead of t=0. The correct approach requires using the known acceleration and the time interval to find the initial velocity. The user acknowledges the mistake and realizes that the previously calculated velocity does not represent the desired time. The conversation highlights the importance of correctly identifying the time frame for kinematic calculations.
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Homework Statement


A car accelerates at 1.95m/s2 along a straight road. It passes two marks that are 28.0m apart at times t=3.95s and t=5.20s. What was the car's velocity at t=0?

Homework Equations

The Attempt at a Solution


t1=3.95s, t2=5.20s, d=28m, t=t12-t1=1.25s, a=1.95m/s2 v0=?
28m=v0*1.25s+1/2*1.95m/s2*(1.25s)^2
v0=21.2 m/s

Could anybody tell me where have I done wrong?
 
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KevinFan said:

Homework Statement


A car accelerates at 1.95m/s2 along a straight road. It passes two marks that are 28.0m apart at times t=3.95s and t=5.20s. What was the car's velocity at t=0?

Homework Equations

The Attempt at a Solution


t1=3.95s, t2=5.20s, d=28m, t=t12-t1=1.25s, a=1.95m/s2 v0=?
28m=v0*1.25s+1/2*1.95m/s2*(1.25s)^2
v0=21.2 m/s

Could anybody tell me where have I done wrong?

You've tried to calculate the velocity at ##t = 3.95s##, although I think you made a mistake in that calculation.
 
PeroK said:
You've tried to calculate the velocity at ##t = 3.95s##, although I think you made a mistake in that calculation.
Thank you, now I see that the velocity I calculated is only the velocity at t= 3.95s:)
 
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