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Two Mark Questions
1. Define the term displacement. Is it a vector quantity or a scalar quantity?
- Displacement is the change in position of an object from its initial point to its final point. It is represented as a straight line connecting these two points and includes both magnitude and direction. Therefore, displacement is a vector quantity.
2. What is circular motion? Is circular motion an accelerated motion?
- Circular motion is the movement of an object along the circumference of a circle. Despite the object having a constant speed, its direction changes continuously, leading to a change in velocity. Since velocity is a vector, a change in direction constitutes acceleration. Therefore, circular motion is an accelerated motion.
3. Derive mathematically the first equation of motion [tex]\( v = u + at \)[/tex].
- Let's define the terms first:
- [tex]\( v \)[/tex] is the final velocity.
- [tex]\( u \)[/tex] is the initial velocity.
- [tex]\( a \)[/tex] is the acceleration.
- [tex]\( t \)[/tex] is the time.
- Acceleration [tex]\( a \)[/tex] is the rate of change of velocity, given by the formula:
[tex]\[ a = \frac{v - u}{t} \][/tex]
- By rearranging this equation to solve for [tex]\( v \)[/tex], we get:
[tex]\[ at = v - u \][/tex]
- Adding [tex]\( u \)[/tex] to both sides, we get:
[tex]\[ v = u + at \][/tex]
- Therefore, the first equation of motion is [tex]\( v = u + at \)[/tex].
Three Mark Questions
4. A boy runs for 20 min. at a uniform speed of 18 km/h. At what speed should he run for the next 40 min. so that the average speed comes 24 km/h?
- First, convert the time to hours:
- [tex]\( t_1 = \frac{20}{60} = \frac{1}{3} \)[/tex] hour for the first part of the run.
- [tex]\( t_2 = \frac{40}{60} = \frac{2}{3} \)[/tex] hour for the second part of the run.
- Calculate the distance covered in the first 20 minutes:
[tex]\[ d_1 = 18 \times \frac{1}{3} = 6 \text{ km} \][/tex]
- The total distance required to maintain an average speed of 24 km/h over 1 hour is:
[tex]\[ \text{Total distance} = 24 \times 1 = 24 \text{ km} \][/tex]
- The distance he needs to cover in the next 40 minutes:
[tex]\[ d_2 = 24 - 6 = 18 \text{ km} \][/tex]
- The speed required for the next 40 minutes:
[tex]\[ \text{Speed} = \frac{d_2}{t_2} = \frac{18}{\frac{2}{3}} = 27 \text{ km/h} \][/tex]
5. A train accelerated from 10 km/h to 40 km/h in 2 minutes. How much distance does it cover in this period?
- Convert the speeds to meters per second (1 km/h = 0.27778 m/s):
- Initial speed [tex]\( u = 10 \times 0.27778 = 2.7778 \text{ m/s} \)[/tex]
- Final speed [tex]\( v = 40 \times 0.27778 = 11.1112 \text{ m/s} \)[/tex]
- Convert the time to seconds:
[tex]\[ t = 2 \times 60 = 120 \text{ seconds} \][/tex]
- Using the equation for uniformly accelerated motion [tex]\( s = \frac{(u + v)}{2} \times t \)[/tex]:
[tex]\[ s = \frac{(2.7778 + 11.1112)}{2} \times 120 = 833.34 \text{ meters} \][/tex]
6. A train starts from rest and accelerates uniformly at the rate of 5 m/s² for 5 sec. Calculate the velocity of the train in 5 sec.
- Given:
- Initial velocity [tex]\( u = 0 \text{ m/s} \)[/tex] (since the train starts from rest)
- Acceleration [tex]\( a = 5 \text{ m/s}^2 \)[/tex]
- Time [tex]\( t = 5 \text{ seconds} \)[/tex]
- Using the first equation of motion [tex]\( v = u + at \)[/tex]:
[tex]\[ v = 0 + (5 \times 5) = 25 \text{ m/s} \][/tex]
- Therefore, the velocity of the train after 5 seconds is 25 m/s.
1. Define the term displacement. Is it a vector quantity or a scalar quantity?
- Displacement is the change in position of an object from its initial point to its final point. It is represented as a straight line connecting these two points and includes both magnitude and direction. Therefore, displacement is a vector quantity.
2. What is circular motion? Is circular motion an accelerated motion?
- Circular motion is the movement of an object along the circumference of a circle. Despite the object having a constant speed, its direction changes continuously, leading to a change in velocity. Since velocity is a vector, a change in direction constitutes acceleration. Therefore, circular motion is an accelerated motion.
3. Derive mathematically the first equation of motion [tex]\( v = u + at \)[/tex].
- Let's define the terms first:
- [tex]\( v \)[/tex] is the final velocity.
- [tex]\( u \)[/tex] is the initial velocity.
- [tex]\( a \)[/tex] is the acceleration.
- [tex]\( t \)[/tex] is the time.
- Acceleration [tex]\( a \)[/tex] is the rate of change of velocity, given by the formula:
[tex]\[ a = \frac{v - u}{t} \][/tex]
- By rearranging this equation to solve for [tex]\( v \)[/tex], we get:
[tex]\[ at = v - u \][/tex]
- Adding [tex]\( u \)[/tex] to both sides, we get:
[tex]\[ v = u + at \][/tex]
- Therefore, the first equation of motion is [tex]\( v = u + at \)[/tex].
Three Mark Questions
4. A boy runs for 20 min. at a uniform speed of 18 km/h. At what speed should he run for the next 40 min. so that the average speed comes 24 km/h?
- First, convert the time to hours:
- [tex]\( t_1 = \frac{20}{60} = \frac{1}{3} \)[/tex] hour for the first part of the run.
- [tex]\( t_2 = \frac{40}{60} = \frac{2}{3} \)[/tex] hour for the second part of the run.
- Calculate the distance covered in the first 20 minutes:
[tex]\[ d_1 = 18 \times \frac{1}{3} = 6 \text{ km} \][/tex]
- The total distance required to maintain an average speed of 24 km/h over 1 hour is:
[tex]\[ \text{Total distance} = 24 \times 1 = 24 \text{ km} \][/tex]
- The distance he needs to cover in the next 40 minutes:
[tex]\[ d_2 = 24 - 6 = 18 \text{ km} \][/tex]
- The speed required for the next 40 minutes:
[tex]\[ \text{Speed} = \frac{d_2}{t_2} = \frac{18}{\frac{2}{3}} = 27 \text{ km/h} \][/tex]
5. A train accelerated from 10 km/h to 40 km/h in 2 minutes. How much distance does it cover in this period?
- Convert the speeds to meters per second (1 km/h = 0.27778 m/s):
- Initial speed [tex]\( u = 10 \times 0.27778 = 2.7778 \text{ m/s} \)[/tex]
- Final speed [tex]\( v = 40 \times 0.27778 = 11.1112 \text{ m/s} \)[/tex]
- Convert the time to seconds:
[tex]\[ t = 2 \times 60 = 120 \text{ seconds} \][/tex]
- Using the equation for uniformly accelerated motion [tex]\( s = \frac{(u + v)}{2} \times t \)[/tex]:
[tex]\[ s = \frac{(2.7778 + 11.1112)}{2} \times 120 = 833.34 \text{ meters} \][/tex]
6. A train starts from rest and accelerates uniformly at the rate of 5 m/s² for 5 sec. Calculate the velocity of the train in 5 sec.
- Given:
- Initial velocity [tex]\( u = 0 \text{ m/s} \)[/tex] (since the train starts from rest)
- Acceleration [tex]\( a = 5 \text{ m/s}^2 \)[/tex]
- Time [tex]\( t = 5 \text{ seconds} \)[/tex]
- Using the first equation of motion [tex]\( v = u + at \)[/tex]:
[tex]\[ v = 0 + (5 \times 5) = 25 \text{ m/s} \][/tex]
- Therefore, the velocity of the train after 5 seconds is 25 m/s.
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