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Tatiana [17]
3 years ago
11

A speedboat increases its speed from 14.5 m/s to 29.3 m/s in a distance of 172 m.

Physics
1 answer:
dalvyx [7]3 years ago
6 0

Answer:

a. Acceleration, a = 1.88 m/s²

b. Time, t = 7.87 seconds.

Explanation:

Given the following data;

Initial velocity, U = 14.5m/s

Final velocity, V = 29.3m/s

Distance, S = 172m

a. To find the acceleration of the speedboat;

We would use the third equation of motion;

V² = U² + 2aS

Substituting into the formula

29.3² = 14.5² + 2a*172

858.49 = 210.25 + 344a

344a = 858.49 - 210.25

344a = 648.24

a = 648.24/344

Acceleration, a = 1.88 m/s²

b. To find the time;

We would use the first equation of motion;

V = U + at

29.3 = 14.5 + 1.88t

1.88t = 29.3 - 14.5

1.88t = 14.8

Time, t = 14.8/1.88

Time, t = 7.87 seconds.

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A stone is dropped into a well. The sound of the splash is heard 3.5 seconds later. What is the depth of the well? Take the spee
Naddika [18.5K]

Answer:

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

s = ut +\frac{1}{2}at^{2}

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

s = \frac{1}{2}gt'^{2}

s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}                     (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

s = v\times t''

s = 343\times t''                                              (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t'                                                                        (3)

Using eqn (2) and (3):

s = 343(3.5 - t')                                                                 (4)

from eqn (1) and (4):

4.9t'^{2} = 343(3.5 - t')

4.9t'^{2} + 343t' - 1200.5 = 0

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

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3 years ago
A police car parked on the side of the highway emits a 1200 Hz sound that bounces off a vehicle farther down the highway and ret
emmasim [6.3K]

f' = frequency observed by the police car after sound reflected from the vehicle and comes back to police car = 1250 Hz

f = frequency emitted by the police car  = 1200 Hz

V = speed of sound = 340 m/s

v = speed of vehicle = ?

frequency observed by the police car is given as

f' = f (V + v)/(V - v)

inserting the values in the above equation

1250 = 1200 (340 + v)/(340 - v)

v = 6.9 m/s

4 0
3 years ago
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