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2 years ago

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A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates

at the rate of 0.500 m/s2 for 7.00 s. (a) What is his final velocity? (b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save? (c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

1 answer:

elena-14-01-66 [18.8K]2 years ago

8 0

Answer:

given,

initial velocity = 11.5 m/s

acceleration = 0.5 m/s²

time = 7 s

a) using equation of motion

v = u + a t

v = 11.5 + 0.5 × 7

v = 15 m/s

b) $s = u t + \dfrac{1}{2}at^2$

$s = 11.5\times 7 + \dfrac{1}{2}\times 0.5\times 7^2$

s = 92.75 m

distance traveled by racer with final velocity

d = 300 - 92.75

= 207.25 m

time spend to travel 207.25

$t = \dfrac{207.25}{15}$

t = 13.82 s

total time spent to travel 300m

t = 7 + 13.82

t = 20.82 s

time required to travel 300 m with initial velocity

$t_0 =\dfrac{d}{v_0}$

$t_0 =\dfrac{300}{11.5}$

= 26.09 s

time saved due to acceleration

t = 26.09 - 20.82 = 5.27 s

c) second racer is 5 m apart from the first one

distance traveled by the second racer till first one finishes the race

$d_2 = v_2t$

$d_2 = 11.8\times 20.82 = 245.676 m$

distance behind the leader

d = 300 -5 -245.676 = 49.34 m

time behind the winner

$t_f = \dfrac{d}{v_2} =\dfrac{49.34}{11.8} = 4.18 s$

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1) The net force is 16 N to the right

2) The net force is 98 N to the left

3) The net force is 0.5 N downward

4) The net force is 170 N to the right

5) The net force is 175 N to the right

Explanation:

1)

To find the net force, we have to analyze all the forces acting on the box.

We have:

Force to the right: $F_a = 20 N$ , the applied force
Force to the left: $F_f = 4 N$ , the force of friction
Force to the bottom: $F_g = 400 N$ , the weight of the box (the weight is always downward vertically)
Force to the top: $F_N = 400 N$ . This is the normal force, which is the reaction force exerted by the table on the box: it points upward and counterbalances the weight of the box, preventing it from falling down)

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While the vertical force is

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So the net force is 16 N to the right.

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In this case, we have the following forces:

$F_g = 4 N$ downward, the weight of the ball
$F_a = 100 N$ to the left, the force that kicks the ball
$F_f = 2 N$ to the right, the force of friction
$F_N = 4 N$ upward, the normal reaction exerted by the field on the ball

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$F_g = 8 N$ downward, the weight of the squirrel
$F_f = 7.5 N$ upward, the air resistance, acting upward

Both forces act vertically and there are no other forces acting in other directions, therefore the net force on the squirrel is simply equal to the net force on the vertical direction, which is:

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While the net force in the vertical direction is

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While the net vertical force is

$F_y = F_N - F_g = 150 - 150 = 0$

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