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Greeley [361]
3 years ago
12

A 7.0-kg rock is subject to a variable force given by the equation f(x)=6.0n−(2.0n/m)x+(6.0n/m2)x2 if the rock initially is at r

est at the origin, find its speed when it has moved 9.0 m.
Physics
1 answer:
stealth61 [152]3 years ago
4 0
For Newton's second law, the force is equal to the product between the mass and the acceleration of the rocket:
F=ma
From which we can rewrite the acceleration as
a(x)= \frac{F}{m} =  \frac{1}{m} (6-2x+6x^2)
where m=7.0 kg.

The velocity of the rocket is the derivative of the acceleration:
v(x) =  \frac{1}{m} (-2+12 x)
and if we substitute x=9.0 m, we find the rocket velocity after 9.0 m:
v(9)= \frac{1}{7}(-2+12\cdot 9)=15.1 m/s

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