A ball is on a horizontal, flat plane at the orgin of your axis system at t0. the ball has an initial velocity of 1.30m/s along the + y axis and, starting at t0 recieves an acceleration of 2.10 m/s^2 in the + x direction. at t= 2.50s a) what is the position of the ball in terms of its x and y coordinates? b) what is the displacement of the ball? c) what is the velocity of the ball?

Respuesta :

It sounds like the ball's velocity along the [tex]y[/tex]-axis should be taken to be constant, in which case the ball has velocity vector

[tex]\mathbf v(t)=\left(2.10\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)\,\hat{\jmath}[/tex]

We note that at [tex]t=t_0[/tex], we're told the ball is at the origin, so its position at this point is [tex]\mathbf r(t_0)=\mathbf 0[/tex]. We then integrate the velocity vector to find the position vector:

[tex]\mathbf r(t)=\mathbf r(t_0)+\displaystyle\int_{\tau=t_0}^{\tau=t}\mathbf v(\tau)\,\mathrm d\tau[/tex]

[tex]\mathbf r(t)=\displaystyle\int_{\tau=t_0}^{\tau=t}\left(2.10\,\dfrac{\mathrm m}{\mathrm s^2}\right)\tau\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)\,\hat{\jmath}\,\mathrm d\tau[/tex]

[tex]\mathbf r(t)=\left(1.05\,\dfrac{\mathrm m}{\mathrm s^2}\right)\tau^2\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)\tau\,\hat{\jmath}\bigg|_{\tau=t_0}^{\tau=t}[/tex]

[tex]\mathbf r(t)=\left(1.05\,\dfrac{\mathrm m}{\mathrm s^2}\right)(t^2-{t_0}^2)\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)(t-t_0)\,\hat{\jmath}[/tex]

For the sake of simplicity, I'll assume [tex]t_0=0[/tex], so that

[tex]\mathbf r(t)=\left(1.05\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)t\,\hat{\jmath}[/tex]

(a) Then at [tex]t=2.50\,\mathrm s[/tex], the ball's position in the plane is given by the vector

[tex]\mathbf r(2.50\,\mathrm m)=(6.56\,\mathrm m)\,\hat{\imath}+(3.25\,\mathrm m)\,\hat{\jmath}[/tex]

which is to say its position in the plane is at the point (6.56, 3.25).

(b) The ball's displacement is the same as its position, but in vector form (as above).

(c) The ball has velocity

[tex]\mathbf v(2.50\,\mathrm s)=\left(5.25\,\dfrac{\mathrm m}{\mathrm s}\right)\,\hat{\imath}+\left(1.3\,\dfrac{\mathrm m}{\mathrm s}\right)\,\hat{\jmath}[/tex]

That is, the ball has a speed of [tex]\|\mathbf v(2.50\,\mathrm s)\|=5.41\,\dfrac{\mathrm m}{\mathrm s}[/tex] and is traveling in a direction of about [tex]13.9^\circ[/tex] relative to the positive [tex]x[/tex]-axis.