Learn Chapter 4-Motion in a Plane with Top Tutors

As a seasoned tutor registered on UrbanPro, let me guide you through this question.

When it comes to rain and motion, understanding relative velocities is key. Here, we have rain falling vertically with a speed of 30 m/s and a woman riding a bicycle with a speed of 10 m/s in the north to south direction.

Now, let's consider the relative motion of the rain with respect to the woman. Since the rain is falling vertically, its relative velocity with respect to the woman would also be vertical, as the woman's horizontal motion does not affect it.

So, despite the woman moving from north to south, the rain still appears to fall vertically relative to her. Therefore, to shield herself from the rain, she should hold her umbrella vertically upwards.

This conceptual understanding is crucial for solving such problems effectively. If you need further clarification or assistance with related topics, feel free to reach out for a session on UrbanPro, where we strive to provide the best online coaching and tuition experience.

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.