Question

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a large toy rocket to the back of a sled and take the modified sled to a large, flat, snowy field. You ignite the rocket and observe that the sled accelerates from rest in the forward direction at a rate of 13.513.5 m/s2 for a time period of 3.503.50 s. After this time period, the rocket engine abruptly shuts off, and the sled subsequently undergoes a constant backward acceleration due to friction of 5.155.15 m/s2. After the rocket turns off, how much time does it take for the sled to come to a stop?
By the time the sled finally comes to a rest, how far has it traveled from its starting point?

Answer 1

1) 9.18 s

In the first part of the motion, the rocket accelerates at a rate of

$a_1=13.5 m/s^2$

For a time period of

$t_1=3.50 s$

So we can calculate the velocity of the rocket after this time period by using the SUVAT equation:

$v_1=u+a_1t_1$

where u = 0 is the initial velocity of the rocket. Substituting a1 and t1,

$v_1=(13.5)(3.50)=47.3 m/s$

In the second part of the motion, the rocket decelerates with a constant acceleration of

$a_2 = -5.15 m/s^2$

Until it comes to a stop, to reach a final velocity of

$v_2 = 0$

So we can use again the same equation

$v_2 = v_1 + a_2 t_2$

where $v_1 = 47.3 m/s$. Solving for t2, we find after how much time the rocket comes to a stop:

$t_2 = -\frac{v_1}{a_2}=-\frac{47.3}{5.15}=9.18 s$

2) 299.9 m

We have to calculate the distance travelled by the rocket in each part of the motion.

The distance travelled in the first part is given by:

$d_1 = ut_1 + \frac{1}{2}a_1 t_1^2$

Using the numbers found in part a),

$d_1 = 0 + \frac{1}{2}(13.5) (3.50)^2=82.7 m$

The distance travelled in the second part of the motion is

$d_2= v_1 t_2 + \frac{1}{2}a_2 t_2^2$

Using the numbers found in part a),

$d_2 = (47.3)(9.18) + \frac{1}{2}(-5.15) (9.18)^2=217.2 m$

So, the total distance travelled by the rocket is

d = 82.7 m + 217.2 m = 299.9 m