An ice sled 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 and the sled moves with 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 s, find the times t1 and t2
t1 + t2 = 90
(1/2)a*t1^2 + a*t1*t2 = 17,500
You know that a = 40. That leaves you with two equations in two unknowns to solve.
To find the times t1 and t2, we can use the equations of motion.
Let's assume:
Acceleration, a = +40 ft/s^2
Initial velocity, u = 0 ft/s
Time the sled accelerates, t1
Time the sled moves with constant velocity, t2
Total distance traveled, s = 17,500 ft
Total time taken, t = 90 s
We can break down the motion into two parts:
1. The sled accelerates from rest to some final velocity v during time t1.
2. The sled moves with a constant velocity v for time t2.
1. Acceleration Phase:
Using the equation of motion, s = ut + 0.5at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time:
17,500 = 0.5 * 40 * t1^2
2. Constant Velocity Phase:
During this phase, the sled maintains a constant velocity, v.
Using the equation of motion, s = vt, where s is the distance traveled and v is the constant velocity:
17,500 = v * t2
We also know that the total time taken is:
t = t1 + t2
Now, let's solve these equations to find t1 and t2.
From equation (1):
17,500 = 20t1^2
Simplifying:
t1^2 = 875
Taking the square root of both sides:
t1 = √875
t1 ≈ 29.58 s
From equation (2):
17,500 = v * t2
We can rearrange it as:
t2 = 17,500 / v
Now, let's solve for v.
Using the equation of motion, v = u + at, where u is the initial velocity, a is the acceleration, and t is the time:
v = 0 + 40 * t1
v = 40 * (t1)
Substituting the value of t1 = √875:
v = 40 * (√875)
Now, substitute this value of v in equation (2):
t2 = 17,500 / (40 * √875)
t2 ≈ 18.63 s
Therefore, t1 ≈ 29.58 s and t2 ≈ 18.63 s.
To find the times t1 and t2, we can use the equations of motion.
Let's break down the motion of the sled into two parts: the acceleration phase (t1) and the constant velocity phase (t2).
During the acceleration phase:
Acceleration (a) = +40 ft/s²
Initial velocity (u) = 0 ft/s (since the sled starts from rest)
Time (t1) = Unknown
Final velocity (v1) = Unknown
We can use the equation of motion: v = u + at, where:
v = final velocity
u = initial velocity
a = acceleration
t = time
For the acceleration phase, we have:
v1 = 0 + 40t1
v1 = 40t1
Now, during the constant velocity phase:
Acceleration (a) = 0 ft/s² (since the sled is moving at a constant velocity)
Initial velocity (u) = v1 (since v1 is the final velocity from the acceleration phase)
Time (t2) = Unknown
Final velocity (v2) = v1 (since the sled maintains a constant velocity)
Using the equation of motion: v = u + at, where a = 0, we have:
v1 = v1 + 0t2
v1 = v1
Since the final velocity (v1) for both phases is the same, we can equate them:
v1 = 40t1 (from the acceleration phase)
v1 = v1 (from the constant velocity phase)
Now, we know that the total distance traveled by the sled is 17,500 ft and the total time is 90 s.
In the acceleration phase (t1), the sled travels the distance covered while accelerating:
Distance = Initial velocity * time + 0.5 * acceleration * time^2
Distance = 0 * t1 + 0.5 * 40 * t1^2
Distance = 20t1^2
In the constant velocity phase (t2), the sled travels the distance covered at constant velocity, which is v1 * t2:
Distance = v1 * t2
Adding up both distances, we get:
20t1^2 + v1 * t2 = 17,500
Now, we have two equations:
v1 = 40t1 (equation 1)
20t1^2 + v1 * t2 = 17,500 (equation 2)
We need to solve these two equations simultaneously to find t1 and t2.
Let's substitute equation 1 into equation 2:
20t1^2 + (40t1) * t2 = 17,500
Simplifying the equation:
20t1^2 + 40t1 * t2 = 17,500
Rearranging the terms:
20t1^2 + 40t1 * t2 - 17,500 = 0
Now, we have a quadratic equation in terms of t1. We can solve this equation to find the values of t1.