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3 years ago

M/s

Physics

1 answer:

SashulF [63]3 years ago

8 0

Answer:

a. Final velocity, V = 2.179 m/s.

b. Final velocity, V = 7.071 m/s.

Explanation:

Given the following data;

Acceleration = 0.500m/s²

a. To find the velocity of the boat after it has traveled 4.75 m

Since it started from rest, initial velocity is equal to 0m/s.

Now, we would use the third equation of motion to find the final velocity.

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.

Substituting into the equation, we have;

$V^{2} = 0^{2} + 2*0.500*4.75$

$V^{2} = 4.75$

Taking the square root, we have;

$V^{2} = \sqrt {4.75}$

Final velocity, V = 2.179 m/s.

b. To find the velocity if the boat has traveled 50 m.

$V^{2} = 0^{2} + 2*0.500*50$

$V^{2} = 50$

Taking the square root, we have;

$V^{2} = \sqrt {50}$

Final velocity, V = 7.071 m/s.

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Its height y is

y = 40 m - 1/2 g t²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 g t²

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v = -g (√(80/g) s)

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Solve y = 0 for t to see how long it's airborne during this bounce:

0 = (14 m/s) t - 1/2 g t²

0 = t (14 m/s - 1/2 g t)

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y = (14 m/s) (5 s - 2.86 s) - 1/2 g (5 s - 2.86 s)²

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(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But y will converge to 0 as t gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

y = (3.5 m/s) t - 1/2 g t²

Solve y = 0 for t :

0 = (3.5 m/s) t - 1/2 g t²

0 = t (3.5 m/s - 1/2 g t)

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As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time t after 5.72 s (but before reaching the ground again) would be

v = 7 m/s - g t

At 6 s, the ball has velocity

(iv) v = 7 m/s - g (6 s - 5.72 s) ≈ 4.26 m/s

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