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laila [671]
2 years ago
12

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?
Physics
1 answer:
pishuonlain [190]2 years ago
4 0

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}

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