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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 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 11.5 m/s^2 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 4.15 m/s^2.

Required:
a. After the rocket turns off, how much time does it take for the sled to come to a stop?
b. By the time the sled finally comes to a rest, how far has it traveled from its starting point?

Sagot :

cjrodriguez89

Answer:

a) t = 9.2s

b) Δx = 242.2 m

Explanation:

a)

  • In order to find the time that the sled traveled since the rocket was turned off, we need to find the first the speed that it had at that moment.
  • Applying the definition of accceleration, since we know that the sled started from rest, we can find the value of the final speed (for this part) as follows:

       [tex]v_{f1} = a_{1} * t_{1} = 11.5m/s2* 3.30 s = 38.0 m/s (1)[/tex]

  • This speed, is just the initial speed for the second part, so we can find the time traveled from the moment the rocket was turned off until it came to an stop, as follows:

       [tex]t_{2} = \frac{v_{f1}}{a_{2} } = \frac{38m/s}{4.15m/s} = 9.2 s (2)[/tex]

b)

  • We need to find find first the displacement when the sled was accelerating.
  • Assuming the acceleration is constant, since it started from rest, we can use the following kinematic equation:

       [tex]v_{f1} ^{2} = 2* a_{1} * x_{1} (3)[/tex]

  • Solving for x₁:

       [tex]x_{1} =\frac{v_{f1}^{2} }{2*a_{1}} =\frac{(38m/s)^{2} }{2*11.5m/s2} =62.8 m (4)[/tex]

  • In the same way, we can use the same equation, replacing the values of the final speed (which becomes zero), initial speed (which is the same as vf1), and a, which becomes -4.15 m/s2 as it is backwards.

       [tex]-v_{f1} ^{2} = 2* a_{2} * x_{2} (5)[/tex]

  • Solving for x₂:

       [tex]x_{2} =\frac{-v_{f1}^{2} }{2*a_{2}} =\frac{-(38m/s)^{2} }{2*(-4.15m/s)^2} =174.0 m (6)[/tex]

  • Δx = x₁ + x₂ = 68.2 m + 174.0 m = 242.2 m (7)
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