The discussion centers on calculating the initial velocity of a car accelerating at 1.95 m/s², which passes two points 28.0 m apart at times t=3.95 s and t=5.20 s. The user initially calculated the velocity at t=3.95 s as 21.2 m/s, but this value represents the velocity at that specific time rather than the initial velocity at t=0. The correct approach involves using the kinematic equation to determine the initial velocity based on the given acceleration and distance.
PREREQUISITESStudents studying physics, particularly those focusing on kinematics, as well as educators seeking to clarify concepts related to acceleration and initial velocity calculations.
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?
Thank you, now I see that the velocity I calculated is only the velocity at t= 3.95s:)PeroK said:You've tried to calculate the velocity at ##t = 3.95s##, although I think you made a mistake in that calculation.