Question

A bicycle travels 9.90 km due east in 0.500 h, then 9.00 km at 15.0° east of north in 0.770 h, and finally another 9.90 km due east in 0.500 h to reach its destination. The time lost in turning is negligible. Assume that east is in the +x-direction and north is in the +y-direction.What is the magnitude of the average velocity for the entire trip? What is the direction of the average velocity for the entire trip?

Answer 1

The average velocity for the entire trip is 17.77 km/h in a direction 15.0° north of east.

Step-by-step explanation:

To find the average velocity for the entire trip, we need to calculate the total displacement and total time taken. Starting with the first leg of the trip, the displacement is 9.90 km east and the time taken is 0.500 h. The velocity for this leg is determined by dividing the displacement by the time, resulting in 9.90 km/0.500 h = 19.80 km/h.

For the second leg, the displacement is 9.00 km at 15.0° east of north. To find the horizontal component of this displacement, we use the formula dx = dcosθ, where d is the displacement magnitude and θ is the angle. Therefore, the horizontal displacement is 9.00 km * cos(15°) = 8.765 km.

The vertical displacement is found using the same formula, but with sin instead of cos: dy = dsinθ, which gives us 9.00 km * sin(15°) = 2.311 km. Adding up the horizontal and vertical displacements gives us a total displacement of 8.765 km east and 2.311 km north. The time taken for this leg is 0.770 h. We can now calculate the velocity by dividing the total displacement magnitude by the time, resulting in sqrt((8.765 km)^2 + (2.311 km)^2)/0.770 h = 11.71 km/h.

For the last leg, the displacement is 9.90 km east and the time taken is 0.500 h. The velocity for this leg is 9.90 km/0.500 h = 19.80 km/h.

To find the average velocity for the entire trip, we sum up the velocities for each leg and divide by the number of legs. In this case, there are three legs, so the average velocity is (19.80 km/h + 11.71 km/h + 19.80 km/h)/3 = 17.77 km/h.

To find the direction of the average velocity, we use the formula tanθ = dy/dx, where θ is the angle and dy and dx are the vertical and horizontal displacements, respectively. Using the values from the second leg of the trip, we have tanθ = 2.311 km/8.765 km, which gives us θ = atan(2.311 km/8.765 km) = 15.0° north of east. Therefore, the direction of the average velocity for the entire trip is 15.0° north of east.

Answer 2

Given data:

* The distance traveled in east in time 0.5 h is 9.9 km

* The distance traveled in east of north in time 0.77 h is 9 km at 15 degree.

* The distance traveled in east in 0.5 h is 9.9 km.

Solution:

The diagramatic representation of the give system is

The net distance traveled in east is,

`d_x=9.9+9\cos (15^(\circ))+9.9`

where 9 cos( 15 degree ) terms is representing the component of 9 distance along the east direction.

Solving the value of total distance in east is,

`d_x=28.49\text{ km}`

The net distance traveled in the north direction is,

`\begin{gathered} d_y=9.9*\cos (90^(\circ))+9*\sin (15^(\circ))+9.9*\cos (90^(\circ)) \\ d_y=0+2.33+0 \\ d_y=2.33\text{ km} \end{gathered}`

The total time taken to complete the trip is,

`\begin{gathered} T=0.5+0.77+0.5 \\ T=1.77\text{ h} \end{gathered}`

The average velocity of bicycle along the east is,

`v_x=(d_x)/(T)`

Substituting the known values,

`\begin{gathered} v_x=(28.49)/(1.77) \\ v_x=16.1kmh^(-1) \end{gathered}`

The average velocity of the bicycle along the north direction is,

`\begin{gathered} v_y=(2.33)/(1.77) \\ v_y=1.32kmh^(-1) \end{gathered}`

Thus, the magnitude of the average velocity is,

`\begin{gathered} v_a=\sqrt[]{16.1^2+1.32^2} \\ v_a=16.15kmh^(-1) \end{gathered}`

The direction of the average velocity is,

`\begin{gathered} \tan (\theta)=(v_y)/(v_x) \\ \tan (\theta)=(1.32)/(16.1) \\ \tan (\theta)=0.082 \\ \theta=4.69^(\circ) \end{gathered}`

Hence, the magnitude of the average velocity is 16.15 km per hour and direction of average velocity is at 4.69 degree east of north.