Question

Initially, a particle is moving at 5.24 m/s at an angle of 35.2° above the horizontal. Four seconds later, its velocity is 6.25 m/s at an angle of 57.5° below the horizontal. What was the particle's average acceleration during these 4.00 seconds in the x-direction (enter first) and the y-direction?

Answer 1

- 0.23

0.56

Step-by-step explanation:

`\underset{v_(o)}{\rightarrow}` = initial velocity of the particle =
`(5.24 Cos35.2) \hat{i} + (5.24 Sin35.2) \hat{j}` =
`(4.28) \hat{i} + (3.02) \hat{j}`

`\underset{v_(f)}{\rightarrow}` = final velocity of the particle =
`(6.25 Cos57.5) \hat{i} + (6.25 Sin57.5) \hat{j}` =
`(3.36) \hat{i} + (5.27) \hat{j}`

t = time interval = 4.00 sec

`\underset{a}{\rightarrow}` = average acceleration = ?

Using the kinematics equation

`\underset{v_(f)}{\rightarrow}` =
`\underset{v_(o)}{\rightarrow}` +
`\underset{a}{\rightarrow}` t

`(3.36) \hat{i} + (5.27) \hat{j}` =
`(4.28) \hat{i} + (3.02) \hat{j}` +
`\underset{a}{\rightarrow}` (4)

`(- 0.92) \hat{i} + (2.25) \hat{j}` = 4
`\underset{a}{\rightarrow}`

`\underset{a}{\rightarrow}` =
`(- 0.23) \hat{i} + (0.56) \hat{j}`

hence

Average acceleration along x-direction = - 0.23

Average acceleration along y-direction = 0.56