an ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +13m/s^2. 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 5.30*10^3 m and the total time is 90s. Find t1,t2, and v.
To solve for t1, t2, and v, we can apply the equations of motion.
1. Let's consider the first phase when the rocket engine is active, from t = 0 to t = t1.
Using the formula for distance covered during acceleration:
s = ut + (1/2)at^2
Since the sled starts from rest, the initial velocity (u) is 0, and the acceleration (a) is +13 m/s^2.
Substituting the known values, we have:
5.30*10^3 = 0 + (1/2)(13)t1^2
10.6*10^3 = 13t1^2
2. Next, let's consider the second phase when the sled moves at a constant velocity v, from t = t1 to t = t2.
During this phase, the distance traveled is given by:
s = vt
Substituting the known total distance traveled and total time, we have:
5.30*10^3 = v(t2 - t1)
Since we know the total time is 90s, we can write:
t2 - t1 = 90 - t1
Substituting into the previous equation:
5.30*10^3 = v(90 - t1)
3. Now we can combine the equations to solve for t1, t2, and v.
Substitute the value of (13t1^2) from equation 1 into equation 2:
10.6*10^3 = v(90 - t1)
Simplifying equation 2 further:
5.30*10^3 = 90v - v(t1)
Rearranging this equation:
v(t1) = 90v - 5.30*10^3
Now, solve for t1:
t1 = (90v - 5.30*10^3) / v
Substitute this expression for t1 into equation 1:
10.6*10^3 = 13((90v - 5.30*10^3) / v)^2
Simplifying further:
10.6*10^3v^2 = 13(90v - 5.30*10^3)^2
Now, we can solve this equation to find the value of v. Simplifying and rearranging this quadratic equation may require a bit of algebraic manipulation.
To find t1, t2, and v, we can use the equations of motion.
First, let's find t1. Since the sled starts from rest and accelerates at +13m/s^2, we need to use the equation for displacement when the acceleration is constant:
s = ut + (1/2)at^2
where s is the total distance traveled, u is the initial velocity (which is 0 since the sled starts from rest), a is the acceleration, and t is the time.
Plugging in the given values:
s = 5.30*10^3 m
a = 13 m/s^2
u = 0
We get:
5.30*10^3 = 0*t1 + (1/2)*(13)*(t1^2)
Simplifying the equation:
5.30*10^3 = 6.5*t1^2
Now, let's find t2. We know that the sled moves with constant velocity v, and the total time is 90s. Since the velocity is constant, the equation for displacement is:
s = vt
Plugging in the given values:
s = 5.30*10^3 m
v = v (as we need to find it)
t = t2
We get:
5.30*10^3 = v*t2
Now, we have two equations:
1) 5.30*10^3 = 6.5*t1^2
2) 5.30*10^3 = v*t2
We can solve these two equations simultaneously to find t1, t2, and v. However, this system of equations does not have unique solutions. We need additional information to determine the values of t1, t2, and v.