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cestrela7 [59]
2 years ago
5

Look at the graph above. It shows how three runners ran a 100 meter race

Physics
2 answers:
Rzqust [24]2 years ago
8 0

Answer:

1. Albert

2. Charlie

3. 05 Seconds

4. 14 seconds

5. 8.33 m/s

Explanation:

The problems in the given scenarios can be solved just by analyzing the given the graph. Here the x-axis of the graph shows the time in seconds and y-axis shows the distance covered.

1. We can see that Albert finished the race in 12 seconds. This is lesser in comparison with Bob (14 seconds) and Charlie (17 seconds). Thus Albert is the winner of the race.

2. We can see that when Charlie reached at 50 meters, the time is increasing while the distance is same. It implies that he stopped there for some rest.

3. This rest was of 05 seconds.

4. Bob completed the race in 14 seconds.

5. Albert covered 100 m in 12 seconds, so his speed (v) will be,

v = \frac{100}{12}  = 8.33 m/s

anastassius [24]2 years ago
7 0

Answer:

1. Albert

2. Charlie

3. 5 seconds

4. 14 seconds

5. 8.33... meters per second

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3 years ago
A baseball is hit almost straight up into the air with a speed of 26 m/s . Estimate how high it goes.
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Answer:

The maximum height of the ball is 34.5 m.

The ball is 5.31 s in the air.

Explanation:

Hi there!

The equations for the height and velocity of the baseball that is hit straight up are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the baseball at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g =  acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t.

If we place the origin of the frame of reference at the place where the baseball is hit, then, y0 = 0.

To calculate how high it goes, we have to obtain the time at which the ball is at maximum height. At that point, the velocity is 0. Then using the equation of velocity:

v = v0 + g · t

0 = 26 m/s - 9.8 m/s² · t

-26 m/s / -9.8 m/s² = t

t = 2.65 s

The height at that time will be the maximum height:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

y = 26 m/s · 2.65 s - 1/2 · 9.8 m/s² · (2.65 s)²

y = 34.5 m

The maximum height of the ball is 34.5 m

If it takes the ball 2.65 s to reach the maximum height it will take another 2.65 s to return to the initial position. Then, the time it will be in the air is (2.65 s + 2.65 s) 5.30 s. However, let´s calculate the time it takes the ball to reach the initial position using the equation for height.

At the initial position y = 0. Then:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

0 = 26 m/s · t - 1/2 · 9.8 m/s² · t²

0 = t (26 m/s - 1/2 · 9.8 m/s² · t)      (t = 0 when the ball is hit)

0 = 26 m/s - 1/2 · 9.8 m/s² · t

-26 / -4.9 m/s² = t

t = 5.31 s     ( the difference with the 5.30 s obtained above is due to rounding the time to 2.65 s).

The ball is 5.31 s in the air.

Have a nice day!

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