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igor_vitrenko [27]
3 years ago
12

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of v0 = 15.0 m/s. The cliff i

s h = 49.0 m above a flat, horizontal beach as shown in the figure. (a) What are the coordinates of the initial position of the stone? x0 = m y0 = m (b) What are the components of the initial velocity? v0x = m/s v0y = m/s (c) Write the equations for the x- and y-components of the velocity of the stone with time. (Use the following as necessary: t. Let the variable t be measured in seconds. Do not include units in your answer.) vx = vy = (d) Write the equations for the position of the stone with time, using the coordinates in the figure. (Use the following as necessary: t. Let the variable t be measured in seconds. Do not state units in your answer.) x = y = (e) How long after being released does the stone strike the beach below the cliff? s (f) With what speed and angle of impact does the stone land? vf = m/s ? =

Physics
1 answer:
ankoles [38]3 years ago
8 0

Answer:

(a) x0 = 0m and y0 = 49.0m

(b) Vox = 15.0m/s Voy = 0m/s

(c) Vx = Vo = 15.0m/s and Vy = -gt

(d) X = 15.0t and y = 49.0 - 4.9t²

(e) t = 3.16s

(f) Vf = 34.4m/s

Explanation:

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While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. how much time does it ta
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While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. The time taken by the car will be 4 seconds. the correct answer is option(c).

When acceleration is constant, the rate of change in velocity is also constant. In the absence of any acceleration, velocity remains constant. When acceleration is positive, velocity becomes more significant.

Let a denote acceleration, u denote initial velocity, v denote final velocity, and t denote time.

The equation of motion is stated as,

v = u + at

v² = u² + 2as

A car travels across a distance of 60.0 m while accelerating constantly from 12.0 m/s to 18.0 m/s.

Then the time taken by the car will be,

u = 12 m/s

v = 18 m/s

s = 60 m

Put these in the equation v² = u² + 2as.

18² = 12² + 2 x a x 60

a = 1.5

Then the time will be

18 = 12 + 1.5t

1.5t = 6

t = 4 seconds

Hence, the time taken is 4 s.

The complete question is:

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. How much time does it take?

1.00 s

2.50 s

4.00 s

4.50 s

To know more about acceleration refer to:  brainly.com/question/12550364

#SPJ4

8 0
1 year ago
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