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## Homework Statement

A 1.25kg object is moving in the x direction at 17.4m/s. Just 3.41s later, it is moving at 26.8m/s at 34.0 degrees to the x-axis. What are the magnitude and direction of the force applied during this time?

## Homework Equations

F=ma

## The Attempt at a Solution

Firstly i drew a diagram and it was in vector components so i calculated the resultant.

1st component

H: 17.4m/s

V: 0m/s

2nd component

H: 26.8cos(34)

V: 26.8sin(34)

Resultant component:

Horizontal = 17.4 + 26.8cos(34) = 39.62 m/s

Vertical = 26.8sin(34)

so [tex]\sqrt{39.62^2+22.22^2}[/tex]

direction Resultant = [tex]tan^{-1}(\frac{22.22}{39.62})[/tex]

direction = 30 degrees

find the magnitude [tex]F = m * \frac{v}{t}[/tex]

[tex]F = 1.25 * \frac{45.43}{3.41}[/tex]

F=16.65N

This is incorrect can someone tell me where i have gone wrong.

P.S