You bicycle 3.2 km east in 0.1 h, then 3.2 km at 15.0 degrees east of north in 0.21 h, and finally another 3.2km due east in 0.1 h to reach its destination. What is the average "velocity" for the entire trip?

Question

You bicycle 3.2 km east in 0.1 h, then 3.2 km at 15.0 degrees east of north in 0.21 h, and finally another 3.2km due east in 0.1 h to reach its destination. What is the average "velocity" for the entire trip?

1 Answer

Impact of this question

5082 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-31T02:51:47

drawn
Let the direction of displacement towards EAST be along positive direction of x-axis and that of NORTH be along positive direction of y-axis. So the unit vector along these directions be $ˆ i and ˆ j$ $hati and hat j$ respectively.

First displacement ( $\to d_{1}$ $vec(d_1$ ) is 3.2 km east in 0.1 h

So

$\to d_{1} = 3.2 ˆ i k m$ $vec(d_1)=3.2hatikm$

Second displacement ( $\to d_{2}$ $vec(d_2$ ) 3.2 km at 15.0 degrees east of north in 0.21 h, This means the direction of displacement makes $75^{\circ}$ $75^@$ with $ˆ i$ $hati$

So

$\to d_{2} = (3.2 {cos 75}^{\circ} ˆ i + 3.2 {sin 75}^{\circ} ˆ j) k m$ $vec(d_2)=(3.2cos75^@hati+3.2sin75^@hatj)km$

Third displacement ( $\to d_{3}$ $vec(d_3$ ) is 3.2 km east in 0.1 h

So

$\to d_{3} = 3.2 ˆ i k m$ $vec(d_3)=3.2hatikm$

Hence net displacement $\to d$ $vec(d)$

$\to d = \to d_{1} + \to d_{2} + \to d_{3}$ $vec(d)=vec(d_1)+vec(d_2)+vec(d_3)$

$= [3.2 ˆ i + (3.2 {cos 75}^{\circ} ˆ i + 3.2 {sin 75}^{\circ} ˆ j) + 3.2 ˆ i] k m$ $=[3.2hati+(3.2cos75^@hati+3.2sin75^@hatj)+3.2hati]km$

$= [6.4 ˆ i + (0.83 ˆ i + 3.09 j)] k m$ $=[6.4hati+(0.83hati+3.09j)]km$

$= [(7.23 ˆ i + 3.09 j)] k m$ $=[(7.23hati+3.09j)]km$

The average velocity

$v_{average} = \frac{∣ ∣ ∣ \to d ∣ ∣ ∣}{Total time} = \frac{\sqrt{{7.23}^{2} + {3.09}^{2}}}{0.1 + 0.21 + 0.1} \frac{k m}{h r}$ $v_"average"=abs(vec(d))/"Total time"=sqrt(7.23^2+3.09^2}/(0.1+0.21+0.1)(km)/(hr)$

$\approx 150.78 km/hr$ $~~150.78"km/hr"$

And the direction of average velocity along North of East is

$θ = {tan}^{- 1} (\frac{3.08}{7.23}) = {23.14}^{\circ}$ $theta =tan^-1(3.08/7.23)=23.14^@$

Science

Math

Humanities

... and beyond

You bicycle 3.2 km east in 0.1 h, then 3.2 km at 15.0 degrees east of north in 0.21 h, and finally another 3.2km due east in 0.1 h to reach its destination. What is the average "velocity" for the entire trip?

1 Answer

Related questions

Impact of this question