An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +35 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 16350 ft and the total time is 90 s.
(a) Find the times t1 and t2.
(b) Find the velocity v.
At the 16350 ft mark, the sled begins to accelerate at -22 ft/s2.
(c) What is the final position of the sled when it comes to rest?
(d) What is the duration of the entire trip?
Start at xo
While accelerating
v = 35 t
d = 0 + 0t + (35/2) t^2
at x1,t1
v1 = 35 t1
x1 = 17.5 t1^2
------------------------
While coasting
v2=v1= 35 t1 + 0
d = 17.5 t1^2 + (35 t1)t + (1/2)0
at x2,t2
x2 = 17.5t1^2 + 35 t1(t2)
so
17.5 t1^2 + 35 t1 t2 =16350
t1+t2 = 90 so t2 = (90-t1)
solve for t1, t2 then x2 etc
To find the times t1 and t2, we can use the equations of motion.
(a) Find t1:
Using the first equation of motion, which relates displacement (s), initial velocity (u), time (t), and acceleration (a):
s = ut + (1/2)at^2
Substituting s = 16350 ft, u = 0 ft/s, and a = 35 ft/s^2, we get:
16350 = 0 + (1/2)(35)t1^2
Simplifying:
32700 = 17.5t1^2
Dividing both sides by 17.5:
t1^2 = 1872
Taking the square root of both sides:
t1 = sqrt(1872) ≈ 43.27 s (rounded to two decimal places)
(b) Find v:
The sled moves with constant velocity v for a time t2. Since the acceleration is zero, we can use the equation:
s = vt
Substituting s = 16350 ft and t = t2, we get:
16350 = v * t2
Simplifying:
v = 16350 / t2
To find t2, we can use the fact that the total time is 90 s:
t1 + t2 = 90
Substituting t1 = 43.27 s, we get:
43.27 + t2 = 90
Solving for t2:
t2 = 90 - 43.27 ≈ 46.73 s (rounded to two decimal places)
Substituting this value of t2 into the equation v = 16350 / t2, we can find the velocity v:
v = 16350 / 46.73 ≈ 349.98 ft/s (rounded to two decimal places)
(c) To find the final position of the sled when it comes to rest, we can use the equation:
s = ut + (1/2)at^2
Since the acceleration is -22 ft/s^2, and the sled comes to rest, the final velocity is 0 ft/s. Thus, we have:
0 = v + (-22)t3
Simplifying:
22t3 = v
Substituting v = 349.98 ft/s, we get:
22t3 = 349.98
Solving for t3:
t3 = 349.98 / 22 ≈ 15.91 s (rounded to two decimal places)
To find the final position, we can use the equation:
s = ut + (1/2)at^2
Substituting u = v (since the sled comes to rest), a = -22 ft/s^2, and t = t3, we get:
s = vt3 + (1/2)(-22)(t3)^2
s = 0 + (-11)(15.91)^2
s ≈ -2828.88 ft (rounded to two decimal places)
The negative sign indicates that the sled has traveled in the opposite direction.
(d) The duration of the entire trip is the sum of t1, t2, and t3:
Duration = t1 + t2 + t3
Substituting the calculated values, we get:
Duration ≈ 43.27 + 46.73 + 15.91
Duration ≈ 105.91 s (rounded to two decimal places)