Car and Ball Acceleration Problems

[Masrat_A]

Maybe I've gotten them figured out, but I'm still very much a physics newbie and thus not feeling confident, so if anyone could please review my work, I'd greatly appreciate it!

1. Homework Statement
1) A car accelerates uniformly from 10 m/s to 20 m/s in 5 seconds.
a) How far does it travel in this time?
b) After it reaches the speed of 20 m/s, how long does it take to stop the car over a distance of 100 m?

2) A ball thrown downward from the top of a 40 m high building takes 2 seconds to hit the floor. Determine its initial and final speeds.

Homework Equations

Please see link below.

The Attempt at a Solution

http://i.imgur.com/iRR84fbh.jpg

 

[andrewkirk]

For 1(a) your logic is correct. However check your arithmetic - what is 50+25?
For 1(b) you have worked out how long it takes for the car to travel 100m is it is not stopping. To calculate the time when it is stopping, assuming constant deceleration, you need to do the same sort of calc as in (a), with initial velocity 20 m/s, final velocity 0m/s and distance traveled =100m. Work out the acceleration a, which will be negative, then work out t.

 

[Masrat_A]

Thank you! I've fixed (a), but for (b), would it be feasible to perform the following?

a = -V₀²/2d = -20²/2(100) = -400/200 = -2 m/s²
t = (V_f - V₀)/a = (0 - 20)/-2 = -20/-2 = 10 sec

Could any of us please provide some assistance with question number two as well?

 

[andrewkirk]

Those calculations look correct.

For Q2, use the equation $d = v_0t+\frac12at^2$. We know the distance $d$, acceleration $a$ and time $t$ so you can work out $v_0$.

 

[Masrat_A]

Could you please tell which is the a precisely? Would it be 20 m/s?

 

[andrewkirk]

$a$ is the acceleration due to gravity, which is $9.8ms^{-2}$.

 

[Masrat_A]

Oh, I see. Thank you again!

 

[Masrat_A]

By following the equation d = v₀t+1/2at², here is what I have gotten:

40 = 2v₀ + (9.8*4)/2
40 = 2v₀ + 10.2
29.8 = 2v₀
14.9 = v₀

Would this be initial speed, or final? If initial, could anyone please explain how we could reach the final or vice-versa?

 

[andrewkirk]

Would this be initial speed, or final?

That equation came from the one you posted in the OP. What does $v_0$ represent in that equation?

Also, there is an arithmetic mistake in the second line of your last post.

 

[Masrat_A]

Whoops, my mistake.

$40 = 2v_0 + (9.8*4)/2$
$40 = 2v_0 + 19.6$
$20.4 = 2v_0$
$10.2 = v_0$

I would say $v_0$ is initial. Does the following look okay for final?

$v_f = v_0 +at$
$v_f = 10.2 + (9.8)(2)$
$v_f = 29.6$

 

[andrewkirk]

Yes, that all looks OK.