A student walked 4km due east and then 15km in a direction of 37 degree north of east, Finally he walked due north for 3km, how much is his displacement?
To find the student's displacement, we need to determine the straight-line distance and direction from the starting point to the ending point.
First, let's break down the given information into its components:
1. The student walks 4 km due east.
2. The student then walks 15 km in a direction of 37 degrees north of east.
3. Finally, the student walks due north for 3 km.
To calculate the displacement, we will sum up the eastward and northward components separately.
Eastward Component:
The student walks 4 km due east, which means there is nothing in the northward direction. Hence, the eastward component is simply 4 km.
Northward Component:
To calculate the northward component, we need to consider both the 15 km northeast (37 degrees north of east) and 3 km north. These distances form a right triangle, and we can use trigonometry to solve for the northward component.
Using the angle of 37 degrees, we can find the northward component as follows:
northward component = 15 km * sin(37°) + 3 km
sin(37°) = 0.6018 (rounded to four decimal places)
northward component = 15 km * 0.6018 + 3 km
northward component = 9.027 km + 3 km
northward component = 12.027 km
Now, we have both the eastward component (4 km) and the northward component (12.027 km). To find the displacement, we will use the Pythagorean theorem:
displacement = √(eastward component^2 + northward component^2)
displacement = √(4 km^2 + 12.027 km^2)
displacement = √(16 km^2 + 144.648 km^2)
displacement = √160.648 km^2
displacement = 12.68 km (rounded to two decimal places)
Therefore, the student's displacement is approximately 12.68 km.