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Kinematics (Don't know what I did wrong )

  • Thread starter BlasterV
  • Start date
A car is sitting motionless at a stop light. The driver notices a truck in his rear view mirror, coming up in the lane beside the car.

When the light turns green, the car begins to accelerate forward at 1.9 meters per second squared. At exactly the same moment, just as the light turns green, the truck reaches the intersection moving with a constant speed 14 meters per second.

How far is the car from the intersection when it catches up to the truck?
How fast is the car going when it catches up to the truck?

Ok. Let me show my work now:
V = 14m/s, A = 1.9m/s^2

I assumed that in order for the car and truck to be at the same spot, You need to match Distances.

So I did: 14m/s * t = 1/2 (1.9m/s^2) t^2
t = 14.74 seconds.
Plug in, And part 1's answer is 206.4 m
Piece of cake, but then it asks the second part.

I really have no idea the approach.
I know I can't use V = AT ( i.e ) V = 14.74 x 1.9
why? because It asks when it catches up to the truck, which is an instantaneous velocity and not an average velocity.

But I'm stuck here. All I know for part 2 of this problem is the car passes the truck at 14.74s

Any help? Thanks
v=at only assumes constant acceleration from zero velocity at t=0. It is instantaneous velocity at time t.
Average velocity would be v= d/t where you know the object was not moving at constant velocity.
Last edited:
Thanks, Got it easily with that, 28m/s. Thanks again :)

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