Question

A proton initially has v=4.0i^−2.0j^+3.0k^ and then 4.0 s later has v=−2.0i^−2.0j^+5.0k^ (in meters per second). For that 4.0 s, what are:

a. The proton’s average acceleration in unit vector notation.
b. The magnitude of a_avg
c. The angle between a_avg and the positive direction of the x axis?

Answer 1

Answer:

$a_{avg}=-1.5i+0.5k$

$1.58113883008\ m/s^2$

$-18.43^{\circ}\ or\ 161.57^{\circ}$

Explanation:

u = 4.0i−2.0j+3.0k v = −2.0i−2.0j+5.0k

Average acceleration is given by

$a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{-2-4, -2+2, 5-3}{4}\\\Rightarrow a=-1.5i-0j+0.5k$

$a_{avg}=-1.5i+0.5k$

The magnitude is

$a_{avg}=\sqrt{(-1.5)^2+0.5^2}\\\Rightarrow a_{avg}=1.58113883008\ m/s^2$

The magnitude is $1.58113883008\ m/s^2$

The angle is

$\theta=tan^{-1}\dfrac{a_z}{a_x}\\\Rightarrow \theta=tan^{-1}\dfrac{0.5}{-1.5}\\\Rightarrow \theta=-18.43^{\circ}\ or\ 161.57^{\circ}$

The angle between $a_{avg}$ and the positive direction of the x axis is $-18.43^{\circ}\ or\ 161.57^{\circ}$