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a puck moves 2.35 m/s in a -22.0 direction. A hockey stick pushes it for 0.125s, changing its velocity to 6.42m/s in a 50.0 direction. what was the magnitude of the accerlation

2 Answers

4 votes

Based on the given parameters, the magnitude of the acceleration is approximately 49.376 m/
s^2.

How to find the magnitude of the acceleration

To find the magnitude of the acceleration, use the equation below;

|a| = |Δv| / Δt

where |Δv| is the magnitude of the change in velocity and Δt is the time interval.

Given:

Initial velocity (vi) = 2.35 m/s at -22.0°

Final velocity (vf) = 6.42 m/s at 50.0°

Time interval (Δt) = 0.125 s

To calculate the change in velocity, subtract the initial velocity vector from the final velocity vector:

Δv = vf - vi

To perform vector subtraction, break down the velocities into their horizontal (x) and vertical (y) components:

Initial velocity components:

vix = vi * cos(-22.0°)

viy = vi * sin(-22.0°)

Final velocity components:

vfx = vf * cos(50.0°)

vfy = vf * sin(50.0°)

Now calculate the change in velocity components:

Δvx = vfx - vix

Δvy = vfy - viy

The magnitude of the change in velocity is given by the Pythagorean theorem:

|Δv| =
\sqrt(\Delta vx^2 + \Delta vy^2)

Finally, calculate the magnitude of the acceleration:

|a| = |Δv| / Δt

Let's compute these values:

vix = 2.35 m/s * cos(-22.0°) ≈ 2.086 m/s

viy = 2.35 m/s * sin(-22.0°) ≈ -0.856 m/s

vfx = 6.42 m/s * cos(50.0°) ≈ 4.100 m/s

vfy = 6.42 m/s * sin(50.0°) ≈ 4.905 m/s

Δvx = 4.100 m/s - 2.086 m/s ≈ 2.014 m/s

Δvy = 4.905 m/s - (-0.856 m/s) ≈ 5.761 m/s

|Δv| =
\sqrt((2.014 m/s)^2 + (5.761 m/s)^2) ≈ 6.172 m/s

|a| = 6.172 m/s / 0.125 s ≈ 49.376 m/
s^2

Therefore, the magnitude of the acceleration is approximately 49.376 m/
s^2.

User Zmey
by
6.6k points
4 votes

Answer:

48.9 m/s²

Step-by-step explanation:

The acceleration is given by ...

a = (change in velocity)/(time)

= ((final velocity) - (initial velocity))/time

= ((6.42∠50° - 2.35∠-22°) m/s)/(0.125 s)

≈ (6.12∠71° m/s)/(0.125 s)

≈ 48.9∠71° m/s²

The magnitude of the average acceleration was 48.9 m/s².

User Colmulhall
by
7.4k points