Need help with a couple of problems.

  • Thread starter Meteo
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You can write this as vi = vf - gt and substitute it into the eqn above.In summary, the first conversation is about a race between a Porsche and a Honda, where the Porsche accelerates faster and therefore wins. The second conversation is about a rocket launching and reaching a certain altitude, and using equations of motion to find the acceleration and velocity of the rocket.
  • #1
Right now I've only learned 3 equations dealing with constant accleration
v_f=v_i+a_s\Delta t
s_f=s_i+v_i\Delta t+1/2a_s(\Delta t)^2[/tex]
v_f^2=v_i^2+2a_s\Delta s

Ok first problem:

A Porsche challenges a Honda to a 400-m race. Because the Porsche's acceleration of 3.5m/s^2 is larger than the Honda's 3.0m/s^2, the Honda gets a 50-m head start. Both cars start accelerating at the same instant.

Who wins?

I used the 3rd equation to get 2800 and then sqrt 2800 to get Vf=53. Then I used the 1st equation to get delta t. Which I got to be about 15.14 sec. For the Honda I did the same thing and got 15.33 sec. So the Porsche wins. I was wondering. Did I do this right?

The second question...
A 1000kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16s, then the motor stops. The rocket altitude 20s after launch is 5100m. You can ignore any effects of air resistance.

This one completely stumped me. Cause no matter which equation I use, I get at least 2 variables... and is the 1000kg even important?
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  • #2
For the first one, you can use the second equation.
I'll indicate p for the posrche and h for the honda.

[tex]s_p=\frac{3.5 \cdot (\Delta t)^2}{2}[/tex]

[tex]s_h= 50 + \frac{3.0 \cdot (\Delta t)^2}{2}[/tex]

Now you could subsitute their final positions being 400m and see which ones needs the least time.

What is the question for problem 2?
  • #3
For problem2.

The rocket lifts off with a high acceleration (I got about 27 m/s²) for the first 16s reaching a velocity of v1. It then travels under gravity effects only for the next 4s reaching a velocity of v2 (v2 =/= 0).

You have,

v0 = 0 at t = 0 and s = 0
v1 = v1 at t = 16 and s = s1
v2 = v2 at t = 20 and s = s1 + s2

where s2 is the distance traveled under gravity effects only.

You can now use your eqns of motion to eliminate v1 (or v2) and the unknown initial accceleration and using the fact that s1 + s2 = 5100m.

I don't know what the question is supposed to be for this question, but knowing v1 and v2, you should now be able to answer it.

Edit: I don't see where the mass is needed for this Q.
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  • #4
s instead of x? him I've never seen that variable change on those motion equations.
  • #5
The questions for problem 2 were find the acceleration of the rocket, then find the velocity as it passes 5100m.

I labeled
t1=16 t2=20

Using the equations I got

I then used the 2nd equation
[tex]5100=128a+16a(4)+1/2(-9.8)4^2 [/tex]
I get a=26.9 :smile:

I don't know where to go from here.
Im not sure how to use s1 + s2 = 5100m.
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  • #6
Meteo said:

I get a=26.9 :smile:

I don't know where to go from here.
Im not sure how to use s1 + s2 = 5100m.
You don't have to go very far from here now.

You've already used s1 and s2, in effect.

s1 = ½a(16)²
s2 = v1(20-16) - g(20 - 16)²
s1 + s2 = ½a(256) + 4v1 - g*4²
5100 =128a + 4(16a) - g*4² since v1 = at1 = 16a

To get the velocity at 5100,

vf = vi - gt

This is the eqn of motion in free fall, under gravity effects only, where vf is the velocity at 5100, vi is the initial velocity (= v1) at the start of its travel under free fall and t is the time taken, during free fall, to reach the height of 5100.

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