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tankabanditka [31]
3 years ago
10

A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853

m/s2. What was the car’s velocity at point “A”?
Physics
1 answer:
Norma-Jean [14]3 years ago
7 0

Recall that

{v_f}^2-{v_i}^2=2a\Delta x

where v_i and v_f are the initial and final velocities, respecitvely; a is the acceleration; and \Delta x is the change in position.

So we have

\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)

\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

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find the x-component of this vector 18.4,0.250. Remember, angles are measured from the +x axis. Find the x-component and y-compo
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