Constant acceleration then constant velocity?

Question

I'm having problems with the one physics problem. Not sure which equation(s) to use. Anyways, here is the problem:
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +43 ft/s2. After some time t1, the rocket engine is shut down and the sled moves with constant velocity v for a time t2. Assume the total distance traveled by the sled is 15950 ft and the total time is 90 s.

I need to find t1, t2, and the velocity.

Answer 1

v = 43 * t1

Distance covered = 0.5*43*t1*t1 + v*t2 = 15950

t1+t2 = 90

Substitute v = 43*t1 and t2 = 90-t1 in second eqn,

We have 43(0.5*t1*t1 + t1*(90-t1)) = 15950

0.5*t1^2 + 90*t1-t1^2 = 370.93

t1^2-180*t1+741.86 = 0

Solving for t1, we have t1 = 4.22 sec

So, v = 43*4.22 ft/s = 181.46 ft/s

t2 = 90-4.22 s = 85.78 s

Answer 2

There is only one variable: t1 (the number of seconds before the engine is shut down). If you know t1, you can calculate any of the other values (t2, total distance traveled, final velocity, etc.).

So write an expression for the distance traveled (in 90 seconds) IN TERMS OF t1. That gives you one equation in one unknown, and you just use algebra to solve it.

I'll go one more step for you. Here's the equation:
Distance = 43 x t1^2/2 + 43 x t1 x (90 - t1) = 15950
(It would help you if you spend time figuring out why that equation represents the situation described in the problem.)

Just simplify the formula and solve the resulting quadratic equation for t1. Then t2 equals 90 - t1, and velocity equals 43 x t1.

Final clue: The average velocity during the 90 seconds is about 177 feet per second (15950 / 90). The sled can accelerate to 177 fps in a little over 4 seconds. So the engine must not run for much more than 4 seconds; otherwise, its average speed would be much too high.

Good luck !

Answer 3

Hi, I see the other answers, but just to explain the easiest way to approach this problem and not have to think so hard about which equation applies.

Draw a Velocity vs. time graph. The graph will start of having a line that begins at the origin, with a slope of 43f/s^2 and then after t1 seconds levels off as a horizontal line which continues for t2 more seconds. The area under a velocity-time curve is distance, so you immediately see that:

triangle + rectangle = distance

1/2 * (43 ft/s^2) * t1^2 + (43 ft/s^2)*t1 *t2 = distance = 15950
you know t1 + t2 = 90

so you can solve 2 equations for 2 variables t1 and t2.
velocity will be the y value of the horizontal part of the graph, or 43 f/s^2 * t1

Answer 4

You have two different phases of motion. Both use the same general equations:

s = u + (vi)t + 1/2 (a)t^2

where s is final position, u is initial position, v is initial velocity, a is acceleration, and t is time (this is the UK version for variables, which uses the acronym SUVAT to help remember how to set up the equation).

vf = vi + at
where vf is final velocity, vi is initial velocity, a is acceleration, and t is time.

Phase 1)
s1=0 + 0 +1/2(43 ft/sec^2) t1^2

v = (43 ft/sec^2) t1

You started at position 0 with an initial velocity of 0, so you only had to worry about acceleration.

Phase 2:
s2 = s1 + v (t2) + 0.

s2 is your final position, which is known to be 15,950 ft. You haven't figured out s1 yet, but you can substitute your equation from Phase 1 in its place (you don't need the zeroes). You don't know v yet either, but you can substitute the equation for it.

Finally, you don't know t2 yet, but you do know it's equal to:
t2 = 90 - t1
So you can substitute (90-t1) into the equation for phase 2.

Now, you've got an equation with only one unknown: t1

15950 = 1/2(43 ft/sec^2) t1^2 + 43(ft/sec^2) (90 - t1)

The rest is doing the algebra to get a quadratic equation which you solve with the quadratic equation.

Answer 5

Here are the equations that apply

x 1 = (1/2) ( 43 ft/s^2) (T1^2) distance travel during T1

V = (43 ft/s^2) (T1) velocity at time = T1

x2 = V(T2-T1) distance traveled at constant velocity

x1 + x2 = 15950 ft

T2 = 90 sec

Solve the math and you're done.

Answer 6

Yes. Both. Depends on the context. Constant velocity occurs when f = ma = P - F = 0 Acceleration occurs when f = ma = P - F > 0. f is the net force acting on a mass m. P is the constant pull or push (force) applied to the object. F is a reaction force, like friction, When P - F = 0, we see that the acceleration a = 0 as well. And that's constant velocity. When P - F > 0, we see that the acceleration a > 0 as well. And that's an accelerating velocity. As asserted, both, it depends. QED.

Answer 7

No. Increasing velocity