Question

A proton initially has V=4.0 i -2.0 j +3.0 k and then 4 seconds later has v= -2.0 i -2.0 j +5.0 k (in meters per second).A) For that 4 seconds, what is the protons average acceleration in unit vector notationB) What is the proton's average acceleration in magnitude and find the angle between the average acceleration vector and the positive direction of the x-axis?

Answer 1

Answer:

Explanation:

$\overrightarrow{u}=4\widehat{i}-2\widehat{j}+3\widehat{k}$

$\overrightarrow{v}=-2\widehat{i}-2\widehat{j}+5\widehat{k}$

time, t = 4 s

(a) Acceleration is defined as the rate of change of velocity.

$\overrightarrow{a}=\frac{\overrightarrow{v}-\overrightarrow{u}}{t}$

$\overrightarrow{a}=\frac{-2\widehat{i}-2\widehat{j}+5\widehat{k}-4\widehat{i}+2\widehat{j}-3\widehat{k}}{4}$

$\overrightarrow{a}=-1.5\widehat{j}+0.5\widehat{k}$

(b) Magnitude of acceleration is

$a = \sqrt{(-1.5)^{2}+(0.5)^{2}}$

a = 1.58 m/s²

It is parallel to x axis.