NationStates Jolt Archive

A car accelerates at 2.0m/s^2 along a straight road. It passes two marks that are 30m apart at times t=4.0s and t=5.0s.

What was the car's initial velocity at t=0s?

i dont get how to answer it if im not given any velocities at t=4 or at t=5

I am sad that I can no longer answer this easily.

I will withhold my answer to save myself from looking like an idiot.

i dont get how to answer it if im not given any velocities at t=4 or at t=5

THINK HARDER.

A car accelerates at 2.0m/s^2 along a straight road. It passes two marks that are 30m apart at times t=4.0s and t=5.0s.

What was the car's initial velocity at t=0s?
Moves 30M in 1 second...eg, vavg over 4-5s=30m/s. You should be able to do it from there.

mmm.. ZERO. ?

Set a velocity, increase or decrease until the difference is 30. Or draw a graph to do it faster.

Please stop starting new threads all the time about your homework. Use your old ones.

would the velocity at t=0 be 22m/s?

would the velocity at t=0 be 22m/s?

I got 21m/s, somebody tell me that's right, and that I haven't smoked away all of the physics I learned long, long ago.

YES! thanks guys i worked it out and it was 21

:fluffle:

A car accelerates at 2.0m/s^2 along a straight road. It passes two marks that are 30m apart at times t=4.0s and t=5.0s.

What was the car's initial velocity at t=0s?

It is traveling about "do your own homework -- you'll be smarter for it" m/s at t=0s.

A rock is tossed straight up with a velocity of + 20m/s When it returns, it falls into a hole 10m deep.

What is the rock's velocity as it hits the bottom of the hole?

i did the vf^2=vi^2+2ad

and got: 24.41, but its seemingly wrong

A rock is tossed straight up with a velocity of + 20m/s When it returns, it falls into a hole 10m deep.

What is the rock's velocity as it hits the bottom of the hole?

i did the vf^2=vi^2+2ad

and got: 24.41, but its seemingly wrong

A student takes a physics class, and is pushed from an open window. Just before he hits the pavement, he is moving at 22 meters per second. How high was the window he was pushed from, and does he have time to calculate the problem in his head before he hits the ground?

A rock is tossed straight up with a velocity of + 20m/s When it returns, it falls into a hole 10m deep.

What is the rock's velocity as it hits the bottom of the hole?

i did the vf^2=vi^2+2ad

and got: 24.41, but its seemingly wrong
-4.9t^2+20t=x, solve for t, (x=-10) then plug t into -9.8t+20. Or you can cheat and use 2a10=v^2, then add 20 m/s to your v. Rather, |2a10=v^2|+20

-4.9t^2+20t=x, solve for t, (x=-10) then plug t into -9.8t+20. Or you can cheat and use 2a10=v^2, then add 20 m/s to your v. Rather, |2a10=v^2|+20

doing that i got vf=34, which is also wrong. thanks very much though

a - acceleration s - distance t - time v1 - initial velocity

Equation of motion - s = v1*t + ½at²

a = 2 m/s² s = 30 m t = 1 s (5 - 4) v1 = ?

30 = v1 * 1 + ½*2*1²

30 = v1 + 1

v1 = 29 m/s

Shqipes, you menioned earlier that you were also interested in calculus. Have you tried relating the position, velocity, and acceleration functions in the context of derivatives and integrals? You may find rectilinear motion to make more sense once you see how the various expressions are derived.

If you've already done that, disregard.

Here's a good one:

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0 seconds later. What was the rockets acceleration? (remember, the bolt leaves the rocket with upward momentum.)

I get an initial rocket acceleration of 5.5 m/(s^2)

Shqipes, you menioned earlier that you were also interested in calculus. Have you tried relating the position, velocity, and acceleration functions in the context of derivatives and integrals? You may rectinlinear motion to make more sense once you see how the various expressions are derived.

If you've already done that, disregard.

Don't overcomplicate this. That's my job. :(