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]
[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]
[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)
