Here's a tough problem for the basics of motion.
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 40 feet/s ^2. After some time T1, the rocket engine is shut down and the sled moves a constant velocity V for a time T2. If the total distance traveled by the sled is 17,500 ft and the total time is 90 seconds, find (A) the times T1 and T2 and (B) the velocity V. At the 17500 ft, tge sled begins to accelerate at -20 ft/s^2. (C) What is the final position of the sled when it comes to rest? (D) How long does it take to come to rest?
Originally posted by RamnedUsing Newton's formulas of motion, I get:
Here's a tough problem for the basics of motion.
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 40 feet/s ^2. After some time T1, the rocket engine is shut down and the sled moves a constant velocity V for a time T2. If the total distance traveled by the sled is 17,500 ft and the total time is 90 seconds, ...[text shortened]... he final position of the sled when it comes to rest? (D) How long does it take to come to rest?
(A) t1 = 4.74 s, t2 = 85.3 s
(B) V = 40*t1 = 189.5 m/s
(C) d(final) = 17500 + 40*t1^2 = 18397.4 m
(D) t3 = 2*t1 = 9.47 s
Originally posted by Ramneds = ut + 0.5at^2
Here's a tough problem for the basics of motion.
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 40 feet/s ^2. After some time T1, the rocket engine is shut down and the sled moves a constant velocity V for a time T2. If the total distance traveled by the sled is 17,500 ft and the total time is 90 seconds, ...[text shortened]... he final position of the sled when it comes to rest? (D) How long does it take to come to rest?
17500 = 0.5(40)(T1^2) + VT2
V = 40(T1)
17500 = 20T1^2 + 40T1T2
T1 + T2 = 90
17,500 = 20T1^2 + 40T1(90 - T1)
20T1^2 - 3600T1 + 17500 = 0
T1^2 - 180T1 + 875 = 0
T1 = (180 +- 170)/2
T1 = 5 or 175
T2 = 85 (or -85)
V = 200
v^2 = u^2 + 2as
0 = 200^2 + 2(-20)s3
s3 = 1000
S = 18,500
v = u + at
0 = 200 - 20T3
T3 = 10
T = 100
Originally posted by SchumiAhhh, right...I used t2 = 90 instead of t2 = 90 - t1. Oops! 😕
s = ut + 0.5at^2
17500 = 0.5(40)(T1^2) + VT2
V = 40(T1)
17500 = 20T1^2 + 40T1T2
T1 + T2 = 90
17,500 = 20T1^2 + 40T1(90 - T1)
20T1^2 - 3600T1 + 17500 = 0
T1^2 - 180T1 + 875 = 0
T1 = (180 +- 170)/2
[b]T1 = 5 or 175
T2 = 85 (or -85)
V = 200
v^2 = u^2 + 2as
0 = 200^2 + 2(-20)s3
s3 = 1000
S = 18,500
v = u + at
0 = 200 - 20T3
T3 = 10
T = 100[/b]
One strategy in a snowball fight is throwing a snowball at a high angle over a flat ground and at the same time, throwing a second one at a low angle timed to arrive before or ath the same time as the first. Assume you do tat, with both snowballs thrown with a speed of 25.0 M/S. The first is thrown at an angle of 70.0 degrees with the horizontal. (A) What angle should the second one be thrown to arrive at the same point as the first? (B) How many seconds later should the second snowball be thrwon after the first in order for both to arrive at the same time?
Originally posted by RamnedLet's see if I can screw up another one!
One strategy in a snowball fight is throwing a snowball at a high angle over a flat ground and at the same time, throwing a second one at a low angle timed to arrive before or ath the same time as the first. Assume you do tat, with both snowballs thrown with a speed of 25.0 M/S. The first is thrown at an angle of 70.0 degrees with the horizontal. (A) What ang ...[text shortened]... ould the second snowball be thrwon after the first in order for both to arrive at the same time?
Vx0 = 25*cos(70)
Vy0 = 25*sin(70)
First we need to find out how long the ball will be in the air, so we start by integrating the acceleration to get the velocity function:
int(dVy/dt)dt = int(a)dt = int(-g)dt
Vy - Vy0 = -gt
t = (Vy - Vy0) / (-g)
Since Vy will be equal and opposite to Vy0 at the point it touches down again, we have:
t = (-Vy0 - Vy0) / (-g) = 2 * Vy0 / g = 2 * 25*sin(70) / 9.81 = 4.79 s
The ball will travel horizontally for this amount of time, so we get:
Dx = Vx0*t = Vx0 * 2 * Vy0 / g = (25*cos(70)) * 2 * (25*sin(70)) / 9.81
Dx = 41.0 m
Now we have to find out which angle will give the same displacement. However, the expression for distance contains the product of cos(70) and sin(70). This is useful, because cos(70) = sin(90-70) = sin(20) and vice versa. Therefore, it will remain unchanged if you sub in an angle of 20 degrees, which is the answer to part (A).
(B) Plugging 20 degrees into the equations above, you get:
t = 2 * (25*sin(20)) / 9.81 = 1.74 s
So you'd have to wait 4.79 - 1.74 = 3.05 s before throwing the second snowball to have them land at the same time.
Originally posted by RamnedIgnoring air resistance, and making various other assumptions, with initial velocity of u and an angle a:
One strategy in a snowball fight is throwing a snowball at a high angle over a flat ground and at the same time, throwing a second one at a low angle timed to arrive before or ath the same time as the first. Assume you do tat, with both snowballs thrown with a speed of 25.0 M/S. The first is thrown at an angle of 70.0 degrees with the horizontal. (A) What ang ...[text shortened]... ould the second snowball be thrwon after the first in order for both to arrive at the same time?
Position = [ut cos(a), ut sin(a) - gt^2/2]
So the snowball reaches the horizontal again at t = 2u sin(a)
At this point the snowball has travelled a horizontal distance
= 2u^2 sin(a)cos(a)/g
= u^2 sin(2a)/g
sin(140) = sin(40)
So if the first snowball is launched at 70 degrees, the second needs to be launched at 20 degrees.
The time difference is 2u/g[sin 70 - sin 20]
Which is 3.05 seconds, give or take any mistakes I just made!
Yes.
Try this one; it helps to have a picture:
Jane has a mass of 50.0 kg and needs to swing across a river to rescue Tarzan. However she must swing into a constant horizontal wind force F on a vine that is initially at an angle of X with the vertical. The horizontal distance of the swing is 50.0 m. The force of the wind F is 110 N. The length of the vine is 40.0 m, and the angle that the vine is initially at (X) is 50.0 degrees. (A) With what minimum speed must Jane begin her swing in order to just make it to the other side? (B) Once the rescue is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing?
Originally posted by RamnedOK. This looks like it needs an energy argument (not least because I started off deriving the equation of motion, and it doesn't look too promising). The minimum take off speed required is the one that leaves them stationary at the other side.
Can nobody get it? Or try it? 😉
It really is difficult for being a problem of motion.
Loss of KE = Work done against wind - Loss of PE
mv^2/2 = FD - mgy
where y is the distance dropped
v^2 = 2Fd/m - 2gy
If the starting angle is a1, and the finishing angle is a2, then the distance 'dropped' is y = L(cos(a2) - cos(a1))
Let's plug in the values.
a1 = 50, and 40sin(a2) = 50 - 40sin(50)
=> a2 = 28.9 degrees
v^2 = 2 x 110 x 50 / 50 - 2 x 9.81 x 40(cos(28.9) - cos(50))
v = 6.12 m/s
I'm pretty confident of the method, although there could easily be a mistake in the details.
Coming back - you haven't provided Tarzan's mass, but the calculation is similar.
Originally posted by mtthwWouldn't the minimum speed be 0 m/s no matter what Tarzan weighs, because the wind is helping instead of hindering on the way back? An undampened (or unimpeded) pendulum will attain the same height at both ends of its swing, so no extra "oomph" is needed when swinging back (with the wind helping) to ensure they reach the original height.
OK. This looks like it needs an energy argument (not least because I started off deriving the equation of motion, and it doesn't look too promising). The minimum take off speed required is the one that leaves them stationary at the other side.
Loss of KE = Work done against wind - Loss of PE
mv^2/2 = FD - mgy
where y is the distance dropped
v^2 = 2Fd ils.
Coming back - you haven't provided Tarzan's mass, but the calculation is similar.
Originally posted by PBE6But here you're not swinging from the same height. Unless I'm misinterpreting the question, the finishing level (coming back) is higher than the starting level.
Wouldn't the minimum speed be 0 m/s no matter what Tarzan weighs, because the wind is helping instead of hindering on the way back? An undampened (or unimpeded) pendulum will attain the same height at both ends of its swing, so no extra "oomph" is needed when swinging back (with the wind helping) to ensure they reach the original height.
If that's so, the force alone won't be enough to get them back.