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1 answer:

Vesna [10]3 years ago

7 0

Complete Question

A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward.

From this motion what is the distance covered

What is the magnitude and direction of the displacement

Answer:

$D = 72 \ m$

Magnitude

$d = - 20 \ m$

Direction

West

Explanation:

From the question we are told that

The first distance covered westward is $d_w_1 = 18 \ m /tex] The first distance covered eastward is [tex]d_e1 = 12 \ m /tex] The second distance covered westward is [tex]d_w_2 = 28 \ m /tex] The second distance covered eastward is [tex]d_e2 = 14 \ m /tex] Generally the distance covered is mathematically represented as [tex]D = d_w1 + d_w2 + d_e1 + d_e2$

=> $D = 18 + 28 + 12 + 14$

=> $D = 72 \ m$

For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative

The magnitude of the displacement is

$d = -d_w1 -d_w2 + d_e1 + d_e2$

=> $d = -18-28 + 12 + 14$

=> $d = - 20 \ m$

The direction is west

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i hope it helps...

6 0

3 years ago

Ben rushin is waiting at a stoplight. when it finally turns green, ben accelerated from rest at a rate of a 6.00 m/s2 for a time

jasenka [17]

In the 4.10 seconds that elapsed, Ben reaches a velocity of

$v_f=v_0+at\implies v_f=0\,\dfrac{\mathrm m}{\mathrm s}+\left(6.00\,\dfrac{\mathrm m}{\mathrm s^2}\right)(4.10\,\mathrm s)$

$\implies v_f=24.6\,\dfrac{\mathrm m}{\mathrm s}$

In this time, his displacement $\Delta x$ satisfies

${v_f}^2-{v_0}^2=2a\Delta x\implies\left(24.6\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(0\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(6.00\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x$