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DanielleElmas [232]

3 years ago

14

A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The ca

r accelerates at a uniform rate of 3.70 m/s^2 and the motorcycle at a uniform rate of 4.40 m/s^2. How much time elapses before the MC overtakes the car? How far will each have traveled during that time?

1 answer:

sergeinik [125]3 years ago

7 0

Answer:

t = 8.45 sec

car distance d = 132.09 m

bike distance d = 157.08 m

Explanation:

GIVEN :

motorcycle is 25 m behind the car , therefore distance need to covered by bike to overtake car is 25+ d, when car reache distance d at time t

for car

by equation of motion

$d = ut + \frac{1}{2}at^2$

u = 0 starting from rest

$d = \frac{1}{2}at^2$

$t^2 = \frac{2d}{a}$

for bike

$d+25 = 0 + \frac{1}{2}*4.40t^2$

$t^2= \frac{d+25}{2.20}$

equating time of both

$\frac{2d}{a} = \frac{d+25}{2.20}$

solving for d we get

d = 132 m

therefore t is $= \sqrt{\frac{2d}{a}}$

$t = \sqrt{\frac{2*132}{3.70}}$

t = 8.45 sec

each travelled in time 8.45 sec as

for car

$d = \frac{1}{2}*3.70 *8.45^2$

d = 132.09 m

fro bike

$d = \frac{1}{2}*4.40 *8.45^2$

d = 157.08 m

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$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

$| \vec{L}_{rod_i} | = r_i * m * v_i$

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

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For our final angular momentum we have:

$| \vec{L}_{rod_f} | = r_f * m * v_f$

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$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

but, as the angular momentum is constant, this must be equal to the initial angular momentum

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