For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velocity vector at time t1, and let v⃗ 2 be the velocity vector at time t2. Consider Δv⃗ =v⃗ 2−v⃗ 1 and Δt=t2−t1. Recall that a⃗ av=Δv⃗ /Δt.
A) For Δt = 0.4 s calculate the magnitude (to four significant figures) of the average acceleration a⃗ av.
B) For Δt = 0.4 s calculate the direction (relative to v⃗ 1) of the average acceleration a⃗ av.
C) For Δt = 0.2 s calculate the magnitude (to four significant figures) of the average acceleration a⃗ av.
D) For Δt = 0.2 s calculate the direction (relative to v⃗ 1) of the average acceleration a⃗ av.
E) For Δt = 7x10^-2 s calculate the magnitude (to four significant figures) of the average acceleration a⃗ av.
F) For Δt = 7x10^-2 s calculate the direction (relative to v⃗ 1) of the average acceleration a⃗ av.

Respuesta :

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

[tex]\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}[/tex]

so that its period is [tex]T=3.644\,\frac{\mathrm s}{\mathrm{rev}}[/tex] (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration [tex]\Delta t[/tex]. Denote by [tex]\theta[/tex] the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

[tex]\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ[/tex]

[tex]\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ[/tex]

[tex]\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ[/tex]

We can then compute the magnitude of the velocity vector differences [tex]\Delta\vec v[/tex] for each time interval by using the law of cosines:

[tex]|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta[/tex]

[tex]\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}[/tex]

and in turn we find the magnitude of the average acceleration vectors to be

[tex]\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}[/tex]

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.

Ver imagen LammettHash