Question

Two forces,

vector F 1 = (5.90î − 5.60ĵ) N
and
vector F 2 = (4.65î − 5.55ĵ) N,
act on a particle of mass 2.10 kg that is initially at rest at coordinates
(−1.75 m, +4.15 m).

(a) What are the components of the particle's velocity at t=10.4s ?
(b) In what direction is the particle moving at t = 10.4 s? (+counterclockwise from the x axis)
(c) What displacement does the particle undergo during the first 10.4 s?
(d) What are the coordinates of the particle at t = 10.4 s? (in x and y meters)

Answer 1

First compute the resultant force F:

`\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N`

`\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N`

`\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N`

Then use Newton's second law to determine the acceleration vector
`\mathbf a` for the particle:

`\mathbf F=m\mathbf a`

`(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a`

`\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)(\rm m)/(\mathrm s^2)`

Let
`\mathbf x(t)` and
`\mathbf v(t)` denote the particle's position and velocity vectors, respectively.

(a) Use the fundamental theorem of calculus. The particle starts at rest, so
`\mathbf v(0)=0`. Then the particle's velocity vector at t = 10.4 s is

`\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^(10)\mathbf a(u)\,\mathrm du`

`\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,(\rm m)/(\mathrm s^2)\right)\bigg|_(u=0)^(u=10.4)`

`\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)(\rm m)/(\rm s)`

If you don't know calculus, then just use the formula,

`v_f=v_i+at`

So, for instance, the velocity vector at t = 10.4 s has x-component

`v_(f,x)=0+\left(5.02(\rm m)/(\mathrm s^2)\right)(10.4\,\mathrm s)=52.2(\rm m)/(\mathrm s^2)`

(b) Compute the angle
`\theta` for
`\mathbf v(10.4\,\mathrm s)`:

`\tan\theta=(-55.2)/(52.2)\implies\theta\approx-46.6^\circ`

so that the particle is moving at an angle of about 313º counterclockwise from the positive x axis.

(c) We can find the velocity at any time t by generalizing the integral in part (a):

`\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du`

`\implies\mathbf v(t)=\left(5.02(\rm m)/(\mathrm s^2)\right)t\,\mathbf i+\left(-5.31(\rm m)/(\mathrm s^2)\right)t\,\mathbf j`

Then using the fundamental theorem of calculus again, we have

`\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^(10.4)\mathbf v(u)\,\mathrm du`

where
`\mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m` is the particle's initial position. So we get

`\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^(10.4)\left(\left(5.02(\rm m)/(\mathrm s^2)\right)u\,\mathbf i+\left(-5.31(\rm m)/(\mathrm s^2)\right)u\,\mathbf j\right)\,\mathrm du`

`\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\frac12\left(\left(5.02(\rm m)/(\mathrm s^2)\right)u^2\,\mathbf i+\left(-5.31(\rm m)/(\mathrm s^2)\right)u^2\,\mathbf j\right)\bigg|_(u=0)^(u=10.4)`

`\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m`

So over the first 10.4 s, the particle is displaced by the vector

`\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m`

or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).

(d) See part (c).