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3 years ago

5

A Jeep accelerated at 2.2 m/s2 until after 18 seconds its displacement was 660 meters. Assuming the Jeep traveled in a straight

line, calculate the initial velocity.

1 answer:

pogonyaev3 years ago

5 0

Ok so we know:
The time (t) is 18seconds
The acceleration (a) is 2.2m/s2
The displacement (r) is 660

Using the equation
$r = ut + \frac{1}{2} a {t}^{2}$
With 'u' being the initial velocity we want, we get:
$660 = 18u + 356.4$
So:
$18u = 303.6$
So:
$u = 16.8666$
So the original/initial velocity was 16.8666 or 16.87 m/s

Hope this helped

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Explanation:

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From the fundamental kinematic equation

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Since the initial velocity is zero hence

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The fundamental kinematic equation is part (a) can also be written as

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The fundamental kinematic equation is part (a) can also be written as

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Also, force exerted on the man is given by $F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}$

Substituting 80 Kg for m, 50 cm which is equivalent to 0.5m for \triangle h and other values as used in part c

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