Question

I only need help with e (bottom of the page).​

Answer 1

Step-by-step explanation:

The box is accelerating along the y-axis at a rate of
`+2.5\:\text{m/s}^2` as well as along the x-axis at a rate of
`+5.1\:\text{m/s}^2.` So the magnitude of the box's total acceleration is given by

`a_T = √(a_x^2 + a_y^2)`

`\:\:\:\:= \sqrt{(5.1\:\text{m/s}^2)^2 + (2.5\:\text{m/s}^2)^2}`

`\:\:\:\:=5.7\:\text{m/s}^2`

The direction of the acceleration
`\theta` with respect to the horizontal direction is given by

`\theta = \tan^(-1)\!\left((a_y)/(a_x)\right) = \tan^(-1)\!\left(\frac{2.5\:\text{m/s}^2}{5.1\:\text{m/s}^2}\right)`

`\:\:\:\:= 26.1°`