Physics-magnitude/ displacement?

[J.live]

Homework Statement

A truck travels 3.45 km north and then makes a 90° left turn and drives another 4.10 km. The whole trip takes 4.50 min.

1)With respect to a two-dimensional coordinate system on the surface of Earth such that the y-axis points north, what is the net displacement vector of the truck for this trip?

2) What is the magnitude of the average velocity for this trip?

The Attempt at a Solution

1) -4.10 i + 3.45j ? Is this right?

2) how will i solve for the magnitude? I mean i know we have to use the distance formula but it didn't work.

If some can explain and show some work in order for me to understand. It will be appreciated

 

[w3390]

So these directions form a 90 degree triangle, where one of the sides is 3.45 km long and the other side is 4.10 km long. However, these are just the components of your displacement. The problem wants to know the distance from start to finish following a straight line. Think A^2+B^2=C^2.

For the second part, you are given the time it takes for the whole trip and you calculated the displacement of the trip. Therefore, you have a distance and a time. What is the relation between velocity, distance, and time?

 

[J.live]

yeah the Pythagorean and theorem.

for the second one, I tried to find the magnitude and divide it by the time but didnt come out right

 

[w3390]

Make sure you make the correct conversions. I don't know what units your answer needs to be in, but it may be km/s, km/hr, or m/s.

 

[rwisz]

1) Your answer for the net displacement vector is correct. The negative sign indicates that the displacement is in the opposite direction of the x-axis.

2) To find the magnitude of the average velocity, we need to first find the total displacement of the truck. This can be done using the Pythagorean theorem:

d = √((-4.10)^2 + (3.45)^2) = 5.30 km

Next, we need to find the total time taken for the trip. Since the truck traveled 3.45 km north and 4.10 km east, we can use the distance formula to find the total distance traveled:

d = √((3.45)^2 + (4.10)^2) = 5.30 km

Now, we can use the equation v = d/t to find the average velocity:

v = (5.30 km) / (4.50 min) = 1.18 km/min

Therefore, the magnitude of the average velocity for this trip is 1.18 km/min.