Hard acceleration problem


by cy19861126
Tags: acceleration
cy19861126
cy19861126 is offline
#1
Oct8-06, 04:44 PM
P: 69
An ice shed powered by a rocket engine starts from rest on a large frozen lake and accelerates at 40 ft/s2. After some time t1, the rocket engine is shut down adn the sled moves with constant velocity v for a time t2. If the total distance traveled by the sled is 17500 ft and the total time is 90 s, find t1 and t2

My work so far... I divided the work into two parts: shed that accelerated and the shed that moves in constant motion. Since I know the equation is d = vi + 0.5at2. I plugged in 0 for vi, since the rocket starts from rest. I plugged in a for 40 ft/s2. Since I don't know what t and d are, I just left them there. I got d1 = 20ti2 Next, since I know d1 + d2 = 17500 ft and t1 + t2 is 90s, I plugged 20ti2 into d1 and got 20ti2 + t2 = 90s. What should I do next??? I tried 3 hours on this hk problem, using different equations. But since I don't know the final velocity, I can't use any other equations for acceleration. I tried to use the value of the total distance, 17500 and total time 90s on equation d = 0.5 (vi + vf)t, but the velocity isn't constant so I can't do anything about it.
Phys.Org News Partner Science news on Phys.org
Lemurs match scent of a friend to sound of her voice
Repeated self-healing now possible in composite materials
'Heartbleed' fix may slow Web performance
radou
radou is offline
#2
Oct8-06, 04:49 PM
HW Helper
radou's Avatar
P: 3,225
I suggest you write down the equations of displacement (i.e. the distance travelled) for every part of the motion first. In the first period, t1, you have constant acceleration, in the second one, t2, you have constant velocity.
cy19861126
cy19861126 is offline
#3
Oct8-06, 06:23 PM
P: 69
Quote Quote by radou
I suggest you write down the equations of displacement (i.e. the distance travelled) for every part of the motion first. In the first period, t1, you have constant acceleration, in the second one, t2, you have constant velocity.
Hi, thanks for the suggestion. I used the equation d = vt and d = at2. For the first part, I plugged in d = at2 and got d1 = 40 * t2. For the second part, I plugged in d = vt, and I got d2 = v(90 - t1) since I don't know the velocity yet, I just put v there. Then I used d1 + d2 = 17500, and plugged in the numbers, and I got (40ft/s2)t1^2 + 90v - t1v = 17500. How can I solve for t1 in this equation? I have two variables in one equation

radou
radou is offline
#4
Oct9-06, 05:38 AM
HW Helper
radou's Avatar
P: 3,225

Hard acceleration problem


Quote Quote by cy19861126
Hi, thanks for the suggestion. I used the equation d = vt and d = at2. For the first part, I plugged in d = at2 and got d1 = 40 * t2. For the second part, I plugged in d = vt, and I got d2 = v(90 - t1) since I don't know the velocity yet, I just put v there. Then I used d1 + d2 = 17500, and plugged in the numbers, and I got (40ft/s2)t1^2 + 90v - t1v = 17500. How can I solve for t1 in this equation? I have two variables in one equation
The equation of displacement for the first part is [tex]d_{1}=\frac{1}{2}at_{1}^2[/tex], and for the second part [tex]d_{2}=vt_{2}[/tex]. Now, the velocity v equals a*t1, so, after plugging that in, you get [tex]d_{2}=at_{1}t_{2}[/tex]. Further on, as you said, you use the condition that d1 + d2 = 17500, which implies [tex]\frac{1}{2}at_{1}^2+at_{1}t_{2}=17500[/tex] (*). Now, since you know that t1+t2 =90, just express t2 from this equation, i.e. t2 = 90-t1, and plug it into the equation marked with (*). Now you can calculate t1 from that equation, and t2 from t2 = 90-t1. I suggest that you convert ft to meters, depending on how your final solution must look like, regarding the units.


Register to reply

Related Discussions
[SOLVED] Forces, Acceleration....Hard Stuff (HELP!) Introductory Physics Homework 2
Switching off alternator during hard acceleration General Engineering 6
HARD problem on mathematical problem and fourier series Math & Science Software 0
A Hard Problem...Need Help Introductory Physics Homework 2
Hard problem Chemistry 5