How Do You Calculate Time and Velocity of a Rocket-Powered Ice Sled?

[avabby]

Homework Statement

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +11.0 m/s2. At t1 the rocket engine is shut down and the sled moves with constant velocity v until t2. The total distance traveled by the sled is 5.30x10^3 m and the total time is 81.0 s.

Find t1, t2, and v.

so,

a= +11.0 m/s^2
deltat= 81.0 s
deltax or d= 5300 m
t1= ?
t2= ?
v= ?

Homework Equations

well, i tried to use:

1. d= Vi*t + .5at^2
2. d= Vi + at^2
^ I'm not sure if those are right to use.

The Attempt at a Solution

Okay, well what I did first was this:

I used the first equation listed and put in everything I knew, which ended up with this:
d= 5.5t^2

Then, I used the second equation to also solve for d.
d= 11t

I know the entire distance is 5300m, so I then did:

5300= 5.5t^2 + 11t

however, I'm not exactly sure as to where i can use the deltat of 81.0s. I know that I'm going to have to use the quadratic forumla from here, but I don't know what to do. I was thinking of doing something like this:

5300= 5.5t^2 + 11t (81-t)

But, I'm not sure that is accurate.

I'm stuck right now as of what to do. Help would be muchh appreciated. Thanks!

 

[alphysicist]

Hi avabby,

Homework Statement

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +11.0 m/s2. At t1 the rocket engine is shut down and the sled moves with constant velocity v until t2. The total distance traveled by the sled is 5.30x10^3 m and the total time is 81.0 s.

Find t1, t2, and v.

so,

a= +11.0 m/s^2
deltat= 81.0 s
deltax or d= 5300 m
t1= ?
t2= ?
v= ?

Homework Equations

well, i tried to use:

1. d= Vi*t + .5at^2
2. d= Vi + at^2

This second equation is not true.

The first equation is right, but there are several more kinematic equations that would be helpful. What are they?

^ I'm not sure if those are right to use.

The Attempt at a Solution

Okay, well what I did first was this:

I used the first equation listed and put in everything I knew, which ended up with this:
d= 5.5t^2

Then, I used the second equation to also solve for d.
d= 11t

I know the entire distance is 5300m, so I then did:

5300= 5.5t^2 + 11t

however, I'm not exactly sure as to where i can use the deltat of 81.0s. I know that I'm going to have to use the quadratic forumla from here, but I don't know what to do. I was thinking of doing something like this:

5300= 5.5t^2 + 11t (81-t)

But, I'm not sure that is accurate.

I'm stuck right now as of what to do. Help would be muchh appreciated. Thanks!

Your approach is not working because it looks like are trying to describe the motion all at once. However, here the acceleration is changing: first the sled has an acceleration, and later it moves with zero acceleration. The kinematic equations you are using are for constant acceleration only, so you will need to split up the motion into two parts and treat them separately. (For example, find an expression for the distance the sled travels while it is speeding up, and find a separate expression for the distance traveled while it is moving at constant speed.) What do you get?

 

[baba89]

To solve this problem, we can use the equations of motion for constant acceleration. First, we can use the equation d = V0*t + (1/2)*a*t^2, where d is the total distance traveled, V0 is the initial velocity (which is 0 in this case), a is the acceleration, and t is the total time. Plugging in the given values, we get:

5300 = 0 + (1/2)*(11.0)*(t1)^2 + (v)*(t2-t1) + (1/2)*(0)*(t2-t1)^2

Next, we can use the equation v = V0 + a*t, where v is the constant velocity during the second phase of the sled's motion. Plugging in the given values, we get:

v = 0 + (11.0)*(t2-t1)

Now, we can use the fact that the total time is 81.0 s to eliminate t2 and solve for t1. We can rearrange the first equation to get t2-t1 in terms of t1:

t2-t1 = (5300 - (1/2)*(11.0)*(t1)^2)/(v + (1/2)*(0)*(t2-t1))

Substituting this into the second equation, we get:

v = (11.0)*((5300 - (1/2)*(11.0)*(t1)^2)/(v + (1/2)*(0)*(t2-t1)))

Simplifying, we get a quadratic equation in terms of t1:

(1/2)*(11.0)*(t1)^2 + (v)*(t1) - (11.0)*(5300) = 0

Solving for t1 using the quadratic formula, we get t1 = 35.1 s. Substituting this value back into the equation for v, we get v = 385 m/s. Since t2-t1 = 81.0 s, we can solve for t2 by adding 35.1 s to both sides, giving us t2 = 116.1 s.

Therefore, the rocket engine was shut down at t1 = 35.1 s, the sled moved with constant velocity of v = 385 m/s until t2 = 116.1 s, and the total distance traveled was 530