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A rocket ship in space is at rest relative to a rock floating in deep space. The
rocket ship accelerates away from the rock at 3 m/s2, while it accelerates it
goes through a displacement of 108 m directly away from the rock. For what
amount of time did it accelerate? Hint: If a variable has the numerical value
zero, then you can substitute the value zero in for that variable before doing
the algebra to make the algebra much simpler.

Respuesta :

KeslerGriffin
So the equation used in this problem is ΔX=V0*T+1/2AT^2 the X is the distance, v0 is initial velocity, T is time, and a is acceleration. So when we plug these values it we get: 108= 0•T+1/2•3•T^2,the 0•t disappears, and the 1/2•3 gets us 1.5, so we have 108=1.5T^2, then we divide 108 by 1.5 which gets us 72=t^2, and we then take the square root and get 8.49=T so the answer is 8.49 seconds.

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