Learn Chapter 4-Motion in a Plane with Top Tutors

As a tutor registered on UrbanPro, I'd approach this problem methodically, ensuring the student understands each step clearly. Firstly, let's break down the given information.

The wind is blowing at 72 km/h towards the North-East direction. This indicates that the wind has two components: one towards the North and the other towards the East.

When the boat is stationary, the flag flutters along the North-East direction due to the wind's effect.

Now, when the boat starts moving at a speed of 51 km/h towards the North, it's important to understand the relative motion between the boat and the wind.

Since the boat is moving North and the wind is also blowing from the North-East, the boat will experience the wind's effect at an angle. This will cause the flag to flutter in a different direction compared to when the boat was stationary.

To find the direction of the flag on the mast of the boat, we can use vector addition to find the resultant direction of the wind relative to the boat's movement. This can be done by adding the vectors representing the boat's velocity and the wind's velocity.

After finding the resultant velocity vector, we can determine the direction it represents, which will give us the direction of the flag on the mast of the boat.

So, to summarize, we need to calculate the resultant velocity vector considering both the boat's velocity towards the North and the wind's velocity towards the North-East, and then determine the direction of this resultant vector, which will give us the direction of the flag on the mast of the boat.

As a seasoned tutor registered on UrbanPro, I'd be delighted to assist you with this physics problem.

Firstly, let's break down the problem. We have a stone tied to the end of a string and whirled in a horizontal circle with constant speed. We're given that the string is 80 cm long and the stone makes 14 revolutions in 25 seconds. We need to find the magnitude and direction of acceleration.

To find the magnitude of acceleration, we can use the formula for centripetal acceleration:

ac=v2rac=rv2

Where:

• acac is the centripetal acceleration,
• vv is the speed of the stone, and
• rr is the radius of the circular path, which is equal to the length of the string.

To find vv, we'll use the formula for speed:

v=2πrTv=T2πr

Where:

• TT is the time taken for one revolution.

Given that the stone makes 14 revolutions in 25 seconds, we can find TT by dividing the total time by the number of revolutions:

T=25 seconds14 revolutionsT=14 revolutions25 seconds

Now, let's calculate TT:

T=2514≈1.79 seconds per revolutionT=1425≈1.79 seconds per revolution

Now, we'll find vv:

v=2π×80 cm1.79 s≈282 cm/sv=1.79 s2π×80 cm≈282 cm/s

Now, let's plug vv and rr into the centripetal acceleration formula:

ac=(282 cm/s)280 cm≈997 cm/s2ac=80 cm(282 cm/s)2≈997 cm/s2

So, the magnitude of the acceleration of the stone is approximately 997 cm/s2997 cm/s2.

As for the direction of the acceleration, it's always directed towards the center of the circular path. This is because centripetal acceleration is the acceleration required to keep an object moving in a circle, and it acts perpendicular to the velocity of the object, towards the center of the circle.

Thus, the direction of the acceleration of the stone is towards the center of the circular path.

In summary, the magnitude of the acceleration of the stone is approximately 997 cm/s2997 cm/s2, and its direction is towards the center of the circular path.