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

Question

3 years ago

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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 13.5 m/s2 for a time period of 3.30 s. After this time period, the rocket engine abruptly shuts off, and the sled subsequently undergoes a constant backward acceleration due to friction of 6.15 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?

Physics

1 answer:

valentina_108 [34] · Answer 1 · 2021-11-18T01:55:54+03:00

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)