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A rocket accelerates from rest at 19 m/s 2 for 12 s. How far will it cover during this time? Answer in units of m
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So the position equation for this type of motion would be \[x=x _{0}+v _{0}t+\frac{ 1 }{ 2 }at ^{2}\] where x_0 is the initial height, v_0 is the initial velocity, a is the acceleration and t is time. Your initial height and velocity in this are zero so those terms drop out leaving you with \[x=\frac{ 1 }{ 2 }at ^{2}\] your acceleration (19m/s^2) and time(12s) are both given in the problem so you should be able to plug them in and solve from here.
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