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?

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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?

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Answer 1 · 2016-12-31T02:51:47

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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

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

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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?

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