Question

At time t=0, constant F begins to act on a rock moving through deep space in the +x direction. For time t>0, which are possible function x(t) for the rock's position : (1)x=4t-3 (2) x=-4t^2+6t-3 (3) x=4t^2+6t-3?

Answer 1

(3). `x=4t^2+6t-3`

Explanation:

Assuming that constant `F` implies constant force, from Newtons second law of motion we know that

`vecF=mveca`
where `m` is mass of rock and `veca` constant acceleration produced due to application of force.

Since force is applied in the direction of `+x` direction, the acceleration produced is also in the same direction.
Fro kinematics we know that distance travelled `x` in time `t` is given by
`x(t)=x_@+ut+1/2at^2` ......(1)
where `x_@` is initial position and `u` is initial velocity. The expression is for `t gt 0`
Comparing with given expressions and noting the differences
(1). `x=4t-3`
Has a missing `t^2` term. Hence not possible.
(2). `x=-4t^2+6t-3`
`t^2` term is `-ve`. Hence not possible.
(3). `x=4t^2+6t-3`
Similar to equation (1). Hence possible.