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.
The formula for calculating velocity at t=0 is v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time.
Velocity at t=0 is the instantaneous velocity at a specific moment in time, while average velocity is the overall velocity over a certain period of time.
Yes, velocity at t=0 can be negative if the object is moving in the opposite direction of its initial velocity.
Acceleration can either increase or decrease the velocity at t=0, depending on the direction of the acceleration relative to the initial velocity.
Calculating velocity at t=0 is important in understanding the motion of an object at a specific moment in time and can help predict its future motion.