A bus travels from west to east with a velocity of 11.5 m/s. A marble rolls on the surface of the bus floor with a velocity of 0.200 m/s north. What is the velocity of the marble relative to the road?

Use vectorial addition to add

Vb=(11.5,0) and Vm=(0,0.200)
The relative velocity of the marble with respect to the road is the sum of the vectors, add up the components to get Vr=(11.5,0.2).
The velocity is the magnitude of the resultant Vr. To determine the magnitude, see the example in:
http://www.jiskha.com/display.cgi?id=1253571803

To find the velocity of the marble relative to the road, we need to combine the velocities of the bus and the marble.

The velocity of the marble relative to the road can be calculated by subtracting the velocity of the bus from the velocity of the marble.

Given:
Velocity of the bus (vb) = 11.5 m/s (east)
Velocity of the marble on the bus (vm) = 0.200 m/s (north)

To find the velocity of the marble relative to the road, we can use vector addition.

Adding the two velocities together, we get:

Velocity of the marble relative to the road (vrelative) = (vb + vm)

Since the velocities are in different directions, we need to be careful with the signs.

The east direction is considered positive, and the north direction is considered positive.

So, adding the two velocities together:

vrelative = vb + vm = 11.5 m/s (east) + 0.200 m/s (north)

However, we need to make sure that the two velocities are expressed in the same direction. Since one is east and the other is north, we need to resolve the north velocity component into an east component.

Since the north velocity is given as 0.200 m/s, we can use trigonometry to find the east component:

north component (y-component) = 0.200 m/s * sin(90°) = 0.200 m/s
east component (x-component) = 0.200 m/s * cos(90°) = 0 m/s

Now we have:

vrelative = 11.5 m/s (east) + 0 m/s (north) = 11.5 m/s (east)

So the velocity of the marble relative to the road is 11.5 m/s to the east.

To find the velocity of the marble relative to the road, we need to consider the velocities of the bus and the marble separately.

The velocity of the bus is given as 11.5 m/s towards the east.

The velocity of the marble is given as 0.200 m/s towards the north.

To find the velocity of the marble relative to the road, we can use vector addition.

Since the velocities of the bus and the marble are in different directions, we need to combine them using vector addition. We can treat these velocities as vectors and use the principles of vector addition to find their resultant velocity.

To do this, we can construct a right-angled triangle. The eastern direction will be the horizontal leg of the triangle and the northern direction will be the vertical leg of the triangle.

Given that the velocity of the bus towards the east is 11.5 m/s and the velocity of the marble towards the north is 0.200 m/s, we can use Pythagoras' theorem to find the magnitude of the resultant velocity:

Resultant velocity = square root of ((11.5 m/s)^2 + (0.200 m/s)^2)

Calculating this, we get:

Resultant velocity = square root of (132.25 m^2/s^2 + 0.0400 m^2/s^2)
= square root of 132.2900 m^2/s^2
= 11.504 m/s

Therefore, the velocity of the marble relative to the road is approximately 11.504 m/s.