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

Answer:

- 0.23

0.56

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