Additional Questions and Answers

1. A car accelerates from rest at a constant rate of 2 m/s² for 10 seconds. How far does the car travel during this time?

Answer:

We can use the equation of motion:

$$s = ut + \frac{1}{2} a t^2$$

Where:

• Initial velocity $u = 0$,
• Acceleration $a = 2 \, \text{m/s}^2$,
• Time $t = 10 \, \text{seconds}$.

Substituting the values:

$$s = 0 + \frac{1}{2} \times 2 \times (10)^2 = \frac{1}{2} \times 2 \times 100 = 100 \, \text{m}$$

The car travels 100 meters during this time.

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2. A ball is thrown horizontally from a height of 50 m. What will be the time taken by the ball to reach the ground and its velocity just before hitting the ground? (Assume $g = 10 \, \text{m/s}^2$)

Answer:

We can use the second equation of motion to find the time taken to reach the ground:

$$s = \frac{1}{2} g t^2$$

Where:

• $s = 50 \, \text{m}$,
• $g = 10 \, \text{m/s}^2$.

Substituting the values:

$$50 = \frac{1}{2} \times 10 \times t^2$$ $$t^2 = \frac{50 \times 2}{10} = 10$$ $$t = \sqrt{10} \approx 3.16 \, \text{seconds}$$

Now, to find the velocity just before hitting the ground, we use the equation:

$$v = u + gt$$

Since the ball is thrown horizontally, $u = 0$:

$$v = 0 + 10 \times 3.16 = 31.6 \, \text{m/s}$$

The time taken to reach the ground is approximately 3.16 seconds, and the velocity just before hitting the ground is 31.6 m/s.

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3. A person walks 3 km north, then 4 km east, and finally 2 km south. What is the total distance traveled and the displacement?

Answer:

• Total Distance Traveled = 3 km + 4 km + 2 km = 9 km
• To find the displacement, we can use the Pythagorean theorem since the movements form a right triangle:
  • The displacement vector will be the hypotenuse of the triangle, where the north-south distance is $3 - 2 = 1 \, \text{km}$ and the east-west distance is 4 km.
  • Displacement = $\sqrt{(4)^2 + (1)^2} = \sqrt{16 + 1} = \sqrt{17} \approx 4.12 \, \text{km}$

Thus, the total distance traveled is 9 km, and the displacement is approximately 4.12 km.

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4. A runner completes 200 m in 25 seconds. What is their average speed? If the runner finishes one lap in a circular path with a radius of 50 m, what is their average velocity?

Answer:

• Average Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{200}{25} = 8 \, \text{m/s}$

To find the average velocity:

• The runner completes one lap in a circular path, so the displacement is zero (because the starting and ending points are the same).
• Average Velocity = 0 m/s

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5. A train is traveling at a speed of 72 km/h. If the brakes are applied and the train decelerates at a rate of 2 m/s², how much time will it take to stop? How far will the train travel before coming to rest?

Answer:

• Initial speed $u = 72 \, \text{km/h} = 20 \, \text{m/s}$,
• Final speed $v = 0 \, \text{m/s}$,
• Acceleration $a = -2 \, \text{m/s}^2$ (negative because it is deceleration).

Using the equation of motion:

$$v = u + at$$ $$0 = 20 + (-2) \times t$$ $$t = \frac{20}{2} = 10 \, \text{seconds}$$

To find the distance:

$$s = ut + \frac{1}{2} a t^2$$ $$s = 20 \times 10 + \frac{1}{2} \times (-2) \times (10)^2 = 200 - 100 = 100 \, \text{m}$$

Thus, it will take 10 seconds for the train to stop, and it will travel 100 meters before coming to rest.

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6. A car is moving at a velocity of 36 km/h. How long will it take to cover a distance of 500 meters?

Answer:

• Initial velocity $v = 36 \, \text{km/h} = 10 \, \text{m/s}$,
• Distance $s = 500 \, \text{m}$.

Using the equation $s = v \times t$:

$$500 = 10 \times t$$ $$t = \frac{500}{10} = 50 \, \text{seconds}$$

Thus, it will take the car 50 seconds to cover the distance of 500 meters.

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7. A stone is dropped from a height of 80 meters. How long will it take to reach the ground? What will be its velocity just before hitting the ground? (Assume $g = 10 \, \text{m/s}^2$)

Answer:

To find the time taken to reach the ground, use the equation:

$$s = \frac{1}{2} g t^2$$

Where:

• $s = 80 \, \text{m}$,
• $g = 10 \, \text{m/s}^2$.

Substituting the values:

$$80 = \frac{1}{2} \times 10 \times t^2$$ $$t^2 = \frac{80 \times 2}{10} = 16$$ $$t = \sqrt{16} = 4 \, \text{seconds}$$

Now, to find the velocity just before hitting the ground, use:

$$v = u + gt$$

Since the stone is dropped, $u = 0$:

$$v = 0 + 10 \times 4 = 40 \, \text{m/s}$$

Thus, the stone will take 4 seconds to reach the ground, and its velocity just before hitting the ground will be 40 m/s.

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8. A cyclist travels around a circular track of radius 100 meters at a constant speed of 15 m/s. What is the cyclist’s velocity and acceleration?

Answer:

• Speed of the cyclist is constant at 15 m/s.
• The cyclist is moving along a circular path, so the direction of motion is constantly changing.

To find the velocity:

• Since the motion is circular, the velocity is always tangent to the circular path. Therefore, the magnitude of the velocity is 15 m/s, but its direction continuously changes.

To find the acceleration:

• The acceleration in uniform circular motion is centripetal acceleration, given by: $$a = \frac{v^2}{r}$$ Where:
• $v = 15 \, \text{m/s}$,
• $r = 100 \, \text{m}$.

Substituting the values:

$$a = \frac{15^2}{100} = \frac{225}{100} = 2.25 \, \text{m/s}^2$$

Thus, the cyclist’s velocity is 15 m/s (tangential to the circle), and the centripetal acceleration is 2.25 m/s².