An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +42 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 15800 ft and the total time is 90 s.
a) find times t1 & t2 (s)
b)find velocity (ft/s)
At the 15800 ft mark, the sled begins to accelerate at -24 ft/s2.
(c) What is the final position of the sled when it comes to rest?
ft
(d) What is the duration of the entire trip?
t1 + t2 = 90
15,800 ft = acceleration distance + coasting distance
(1/2)*42*t1^2 + (42*t1)*t2 = 15800
Solve those two equations in the two unknowns, t1 and t2. The substitution method should work.
(b) V = 42 * t1 (after acceleration)
Use the value of t1 derived in part (a) to compute the initial V when deceleration begins.
(c) Let t' = deceleration time
24 t' = V
Solve for t'
Additional distance travelled while decelerating = (V/2)*t'
Add that to 15,800 for final position.
jewish!!!
(a) As the sled accelerates at +42 ft/s^2 for some time t1 and then moves with constant velocity v for a time t2, the total distance traveled can be expressed as:
Distance = (1/2)(a)(t1)^2 + v(t1) + (v)(t2)
Given that the total distance is 15800 ft and the total time is 90 s, we can set up the following equation:
15800 = (1/2)(42)(t1)^2 + 42t1 + vt2 (Equation 1)
(b) Since the sled moves with constant velocity v, we can write:
v = s / t2
where s is the distance covered during the constant velocity motion. As given in the problem, the sled covers a distance of 15800 ft, so we can express v as:
v = 15800 / t2 (Equation 2)
(c) At the 15800 ft mark, the sled starts decelerating at -24 ft/s^2. We can determine the time it takes for the sled to come to rest using the formula:
vf^2 = vi^2 + 2as
Where vf is the final velocity, vi is the initial velocity, a is acceleration, and s is the distance covered. Rearranging the equation, we get:
vf^2 = vi^2 - 2as
Since the final velocity when the sled comes to rest is 0 ft/s, we have:
0 = v^2 - 2(-24)(15800)
Solving for v^2, we find:
v^2 = 2(24)(15800)
v^2 = 758400
Taking the square root, we get:
v = sqrt(758400) ft/s
(d) The duration of the entire trip can be found by adding the times t1 and t2:
Total time = t1 + t2
Let's solve these equations now.
(Note: The calculations will be serious, and not humorous, as this is a complex physics problem.)
Using Equation 2, we can express v in terms of t2:
v = 15800 / t2 (Equation 2)
From Equation 1, we substitute the expression for v from Equation 2:
15800 = (1/2)(42)(t1)^2 + 42t1 + (15800 / t2)(t2)
15800 = (1/2)(42)(t1)^2 + 42t1 + 15800
Since the first term on the right side cancels out:
0 = (1/2)(42)(t1)^2 + 42t1
Now, we can solve this quadratic equation:
0 = 21(t1)^2 + 42t1
0 = t1(21t1 + 42)
Using the zero-product property, we have two solutions:
t1 = 0 (t1 cannot be zero as the sled accelerates at some time)
21t1 + 42 = 0
Solving the second equation:
21t1 + 42 = 0
21t1 = -42
t1 = -42 / 21
t1 = -2 s
Since time cannot be negative, we discard t1 = -2 s.
So, the time t1 is 2 seconds.
To find the time t2, we substitute the value of t1 in Equation 2:
v = 15800 / t2
sqrt(758400) = 15800 / t2
Squaring both sides:
758400 = (15800 / t2)^2
758400 = (15800)^2 / (t2)^2
Multiplying both sides by (t2)^2:
758400(t2)^2 = (15800)^2
Dividing both sides by 758400:
(t2)^2 = (15800)^2 / 758400
Taking the square root:
t2 = sqrt((15800)^2 / 758400)
t2 ≈ sqrt(249364000 / 758400)
t2 ≈ sqrt(328.6462)
t2 ≈ 18.12 s
Therefore, the time t2 is approximately 18.12 seconds.
(b) To find the velocity v, we can substitute the values of t1 and t2 into Equation 2:
v = 15800 / t2
v = 15800 / 18.12
v ≈ 871.46 ft/s
Therefore, the velocity is approximately 871.46 ft/s.
(c) Since the final position when the sled comes to rest is not specified, we cannot determine the exact final position without more information.
(d) The duration of the entire trip is the sum of t1 and t2:
Total time = t1 + t2
Total time ≈ 2 + 18.12
Total time ≈ 20.12 s
Therefore, the duration of the entire trip is approximately 20.12 seconds.
Let's break down the problem into different scenarios and solve them step-by-step:
a) Finding times t1 and t2:
1. The sled starts from rest and accelerates at +42 ft/s^2 until the rocket engine is shut down. Let's assume it accelerates for time t1.
2. After the rocket engine is shut down, the sled moves with constant velocity v for time t2.
3. The total distance traveled by the sled is 15800 ft, and the total time is 90 s.
4. Using the first equation of motion: distance = initial velocity × time + (1/2) × acceleration × time^2, we can set up the equation for the first scenario:
15800 = 0 × t1 + (1/2) × 42 × t1^2
Rearranging the equation, we get:
21t1^2 = 15800
Dividing both sides by 21, we have:
t1^2 = 15800 / 21
Taking the square root of both sides, we find:
t1 ≈ √(15800 / 21)
t1 ≈ √(752.38)
t1 ≈ 27.42 seconds
Now, to find t2, we subtract t1 from the total time:
t2 = 90 - 27.42
t2 ≈ 62.58 seconds
Therefore, the times t1 and t2 are approximately 27.42 seconds and 62.58 seconds, respectively.
b) Finding velocity:
1. The sled moves with constant velocity v for time t2.
2. Using the equation of motion: velocity = initial velocity + acceleration × time, we can set up the equation for the second scenario:
v = 0 + 42 × t2
Substituting the value of t2 we found earlier:
v = 0 + 42 × 62.58
v ≈ 2632.36 ft/s
Therefore, the velocity v is approximately 2632.36 ft/s.
c) Finding the final position of the sled when it comes to rest:
1. At the 15800 ft mark, the sled begins to decelerate at -24 ft/s^2 until it comes to rest.
2. Let's assume the time taken for the sled to come to rest is t3.
3. Using the equation of motion: velocity = initial velocity + acceleration × time, we can set up the equation for the deceleration scenario:
0 = v + (-24) × t3
Substituting the value of v we found earlier:
0 = 2632.36 + (-24) × t3
Solving for t3:
24t3 = -2632.36
t3 ≈ -109.68 seconds (negative value is due to deceleration)
Since time cannot be negative, we discard this result and consider that the sled never comes to rest at the 15800 ft mark.
d) Finding the duration of the entire trip:
The duration of the entire trip is the sum of t1, t2, and t3 (if applicable):
Duration = t1 + t2 + t3 (if applicable)
Duration ≈ 27.42 + 62.58 + 0 (since t3 is not applicable in this case)
Duration ≈ 90 seconds
Therefore, the duration of the entire trip is approximately 90 seconds.
To solve this problem, we will use the equations of motion to find the values of t1, t2, v (velocity), and the final position of the sled.
a) Finding times t1 & t2:
Let's assume the time t1 is the time it takes for the sled to accelerate with the rocket engine.
We know that the acceleration (a) is +42 ft/s^2, and the initial velocity (u) is 0 ft/s.
Using the formula: distance (d) = initial velocity (u) * time (t) + 0.5 * acceleration (a) * time (t)^2.
Since the sled starts from rest, the distance traveled during the time t1 is:
d = 0.5 * 42 * t1^2.
Now, let's consider the remaining time t2, where the sled moves with constant velocity v.
Since the sled moves with a constant velocity, we can use the formula:
v = u + a * t2.
We know that the total distance traveled by the sled is 15800 ft and the total time is 90 s.
Using this information, we can express the total distance traveled as the sum of the distance traveled during time t1 and time t2:
15800 = 0.5 * 42 * t1^2 + v * t2.
Now, we have two equations:
1. d = 0.5 * 42 * t1^2.
2. 15800 = 0.5 * 42 * t1^2 + v * t2.
To find t1 and t2, we need to solve these two equations simultaneously.
b) Finding velocity (v):
Given that the sled moves with constant velocity v during time t2, we can find v using the equation:
v = (15800 - 0.5 * 42 * t1^2) / t2.
c) Finding the final position of the sled when it comes to rest:
At the 15800 ft mark, the sled begins to decelerate at -24 ft/s^2.
Using the formula: final velocity (v) = initial velocity (u) + acceleration (a) * time (t).
The final velocity (v) is 0 ft/s, the initial velocity (u) is v, and the acceleration (a) is -24 ft/s^2.
We can solve for the time (t) using the equation:
0 = v - 24t.
When the sled comes to rest, the final position is given by:
final position = initial position + initial velocity * time + (0.5 * acceleration * time^2).
d) Finding the duration of the entire trip:
The duration of the entire trip is the sum of times t1, t2, and the time it takes for the sled to come to rest at the 15800 ft mark.