Answer:
a) v1 = a1*t1 = 2162 m/s
b) v2 = v1 + a2*(t2-t1) = 2591 m/s
c) [tex]Dt = D1 + D2 = \frac{v1^{2}}{2*a1} + \frac{v2^{2}-v1^{2}}{2*a2}=51004.5m[/tex]
Explanation:
For any movement with constant acceleration:
Vf = vo + a*Δt. Replacing the propper values, with get the answers for parts a) and b):
v1 = a1*t1 = 2162 m/s
v2 = v1 + a2*(t2-t1) = 2591 m/s
Using the formula for displacement we can calculate the total distance asked on part c):
[tex]V_{f}^{2}=V_{o}^2+2*a*D[/tex] Solving for D and replacing the values for each part of the launch:
[tex]D=\frac{V_{f}^{2}-V_{o}^{2}}{2*a}[/tex]
D1 = 24863m
D2 = 26141.5m
Finally we add D1 + D2 for the total distance:
D = 51004.5m