Question

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. If the total distance traveled by the sled is 16350 ft and the total time is 90 s. Find the following. (a) Find the times t1 and t2. t1 t2 (b) Find the velocity v. ft/s At the 15750 ft mark, the sled begins to accelerate at -21 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?

Answer 1

(a) 5.35 s, 84.65 s

The distance covered by the rocket during the first part of the motion (accelerated motion) is

$d_1 = \frac{1}{2}at_1 ^2$ (1)

where

a = +35 ft/s^2 is the acceleration

While the distance covered during the second part of the motion (uniform motion) is

$d_2 = v t_2$

where v is the constant speed reached after the first part of the motion, which is given by

$v=at_1$

So we can rewrite the equation as

$d_2 = a t_1 t_2$ (2)

We also know:

$d= d_1 + d_2 = 16350$ (3) is the total distance

$t= t_1 +t_2 = 90$ (4) is the total time

We can rewrite (3) as

$\frac{1}{2}at_1^2 + a t_1 t_2 = 16350$ (5)

And (4) as

$t_2 = 90-t_1$

Substituting into (5), we get

$\frac{1}{2}at_1^2 + a t_1 (90-t_1) = 16350\\0.5 at_1^2 + 90at_1 - at_1^2 = 16350\\-0.5at_1^2 +90at_1 = 16350$

Substituting a=+35 ft/s^2, we have

$-17.5 t_1^2 +3150 t_1 - 16350 = 0$

which has two solutions:

$t_1 = 174.65 s$ --> discarded, since the total time cannot be greater than 90 s

$t_1 = 5.35 s$ --> this is the correct solution

And therefore $t_2$ is

$t_2 = 90-t_1 = 90 -5.35 s=84.65 s$

(b) 187.25 ft/s

The velocity v is the velocity at the end of the accelerated motion of the rocket, so after

$t_1 = 5.35 s$

The acceleration is

a = +35 ft/s^2

Therefore, the velocity after the first part is:

$v=at_1 = (+35 ft/s^2)(5.35 s)=187.25 ft/s$

(c) 16,585 ft

At the 15750 ft mark, the velocity of the sled is

$v=187.25 ft/s$

Then, it starts decelerating with acceleration

a = -21 ft/s^2

So, its distance covered during this part of the motion will be given by:

$v_f ^ 2 -v^2 = 2ad_3$

where

vf = 0 is the final speed

v = 187.25 ft/s is the initial speed

a = -21 ft/s^2 is the acceleration

d3 is the distance covered during this part

Solving for d3,

$d_3 = \frac{v_f^2 -v^2}{2a}=\frac{0-(187.25 ft/s)^2}{2(-21 ft/s^2)}=835 ft$

So, the final position of the sled will be

$x_3 = 15750 ft + 835 ft =16,585 ft$

(d) 93.92 s

The duration of the first part of the motion is

$t_1 = 5.35 s$

The distance covered during this part is

$d_1 = \frac{1}{2}at_1^2=\frac{1}{2}(+35 ft/s^2)(5.35 s)^2=501 ft$

In the second part (uniform motion), the sled continues with constant speed until the 15750 ft mark, so the distance covered is

$d_2 = 15750- 835 =14915$

And so the duration of the second part is

$t_2 =\frac{d_2}{v}=\frac{14915}{187.25}=79.65 s$

While the duration of the third part (decelerated motion) is

$a=\frac{v_f-v}{t_3}\\t_3 = \frac{v_f-f}{a}=\frac{0-(187.25 ft/s)}{-21 ft/s^2}=8.92 s$

So the total duration is

t = 5.35 s+79.65 s+8.92 s=93.92 s