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14 days ago

14

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

e 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?

1 answer:

Maru [1K]14 days ago

6 0

1) 9.18 s

During the initial phase, the rocket accelerates at

$a_1=13.5 m/s^2$

over a time interval of

$t_1=3.50 s$

The final velocity after this period is found using the SUVAT formula:

$v_1=u+a_1t_1$

with initial velocity u = 0. Substituting a1 and t1 gives:

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

Afterward, the rocket decelerates uniformly at

$a_2 = -5.15 m/s^2$

until it stops, meaning the final velocity is

$v_2 = 0$

Again using the SUVAT formula,

$v_2 = v_1 + a_2 t_2$

and knowing $v_1 = 47.3 m/s$ , solve for t2, the time until the rocket halts:

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

2) 299.9 m

Calculate distances covered in both phases.

Distance in the first phase:

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

Substituting values from part 1,

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

Distance in the deceleration phase:

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

Substituting known values,

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

Total distance traveled is

d = 82.7 m + 217.2 m = 299.9 m

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