A man runs 300 m West in 60 seconds. He then runs 100 m
North-west in 20 seconds
To find the man's overall displacement and direction, we can break down his two movements into their respective horizontal and vertical components and then use vector addition to determine the resultant displacement.
First, let's calculate the horizontal and vertical components of the man's first movement:
Horizontal component: The man runs 300 m west. Since west is in the negative x-direction, the horizontal component of this movement is -300 m.
Vertical component: The man doesn't move vertically during his first movement, so the vertical component is 0 m.
Next, let's calculate the horizontal and vertical components of the man's second movement:
Horizontal component: The man runs 100 m northwest. To find the horizontal component, we need to determine the magnitude and direction of the northwest direction. Since northwest is at a 45-degree angle between west and north, we can use trigonometry to calculate the horizontal component. The magnitude of the northwest direction can be found using the Pythagorean theorem:
magnitude = sqrt(100^2 + 100^2) = sqrt(20000) ≈ 141.4 m
Since northwest is at a 45-degree angle, its horizontal component is equal to its vertical component. Therefore, the horizontal component is 141.4 m in the negative x-direction.
Vertical component: Similar to the horizontal component, the vertical component of the northwest movement is equal to its horizontal component, which is -141.4 m.
Now, let's find the resultant displacement by adding the horizontal and vertical components of both movements:
Horizontal component total = -300 m + (-141.4 m) = -441.4 m
Vertical component total = 0 m + (-141.4 m) = -141.4 m
To find the magnitude of the resultant displacement, we can use the Pythagorean theorem:
magnitude = sqrt((-441.4 m)^2 + (-141.4 m)^2) ≈ 472.7 m
Finally, to determine the direction of the resultant displacement, we can use trigonometry:
angle = arctan((-141.4 m)/(-441.4 m)) ≈ 17.17 degrees
Therefore, the man's overall displacement is approximately 472.7 m at an angle of 17.17 degrees south of west.