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 = -(v_1)/(a_2)=-(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 + (1)/(2)a_1 t_1^2`

Using the numbers found in part a),

`d_1 = 0 + (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 + (1)/(2)a_2 t_2^2`

Using the numbers found in part a),

`d_2 = (47.3)(9.18) + (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