3. express each of the following in m s⁻¹ :
(a) 10 km h⁻¹ (b) 18 km min⁻¹
4. arrange the following speeds in increasing order :
10 m s⁻¹, 1 km min⁻¹, 18 km h⁻¹.
5. a train takes 3 h to travel from agra to delhi with a uniform speed of 65 km h⁻¹. find the distance between the two cities.
6. a car travels first 30 km with a uniform speed of 60 km h⁻¹ and then next 30 km with a uniform speed of 40 km h⁻¹. calculate : (i) the total time of journey, (ii) the average speed of the car.
7. a train takes 2 h to reach station b from station a, and then 3 h to return from station b to station a. the distance between the two stations is 200 km. find : (i) the average speed, (ii) the average velocity of the train.
8. a car moving on a straight path covers a distance of 1 km due east in 100 s. what is (i) the speed and (ii) the velocity, of car ?
9. a body starts from rest and acquires a velocity 10 m s⁻¹ in 2 s. find the acceleration.

Question

3. express each of the following in m s⁻¹ :
(a) 10 km h⁻¹  (b) 18 km min⁻¹
4. arrange the following speeds in increasing order :
10 m s⁻¹, 1 km min⁻¹, 18 km h⁻¹.
5. a train takes 3 h to travel from agra to delhi with a uniform speed of 65 km h⁻¹. find the distance between the two cities.
6. a car travels first 30 km with a uniform speed of 60 km h⁻¹ and then next 30 km with a uniform speed of 40 km h⁻¹. calculate : (i) the total time of journey, (ii) the average speed of the car.
7. a train takes 2 h to reach station b from station a, and then 3 h to return from station b to station a. the distance between the two stations is 200 km. find : (i) the average speed, (ii) the average velocity of the train.
8. a car moving on a straight path covers a distance of 1 km due east in 100 s. what is (i) the speed and (ii) the velocity, of car ?
9. a body starts from rest and acquires a velocity 10 m s⁻¹ in 2 s. find the acceleration.

Accepted Answer

### Question 3 (a)
# Explanation:
## Step1: Convert km to m and h to s
We know that $ 1 \text{ km} = 1000 \text{ m} $ and $ 1 \text{ h} = 3600 \text{ s} $. So, $ 10 \text{ km h}^{-1} = 10 \times \frac{1000 \text{ m}}{3600 \text{ s}} $
## Step2: Calculate the value
$ 10 \times \frac{1000}{3600} = \frac{10000}{3600} \approx 2.78 \text{ m s}^{-1} $
# Answer:
$ \approx 2.78 \text{ m s}^{-1} $

### Question 3 (b)
# Explanation:
## Step1: Convert km to m and min to s
We know that $ 1 \text{ km} = 1000 \text{ m} $ and $ 1 \text{ min} = 60 \text{ s} $. So, $ 18 \text{ km min}^{-1} = 18 \times \frac{1000 \text{ m}}{60 \text{ s}} $
## Step2: Calculate the value
$ 18 \times \frac{1000}{60} = 18 \times \frac{50}{3} = 300 \text{ m s}^{-1} $
# Answer:
$ 300 \text{ m s}^{-1} $

### Question 4
# Explanation:
## Step1: Convert all speeds to $ \text{m s}^{-1} $
- For $ 10 \text{ m s}^{-1} $, it is already in $ \text{m s}^{-1} $.
- For $ 1 \text{ km min}^{-1} $: $ 1 \text{ km} = 1000 \text{ m} $, $ 1 \text{ min} = 60 \text{ s} $, so $ 1 \times \frac{1000 \text{ m}}{60 \text{ s}} \approx 16.67 \text{ m s}^{-1} $
- For $ 18 \text{ km h}^{-1} $: $ 18 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 5 \text{ m s}^{-1} $
## Step2: Arrange in increasing order
Now we have speeds: $ 5 \text{ m s}^{-1} $ (18 km h⁻¹), $ 10 \text{ m s}^{-1} $, $ 16.67 \text{ m s}^{-1} $ (1 km min⁻¹)
# Answer:
$ 18 \text{ km h}^{-1} < 10 \text{ m s}^{-1} < 1 \text{ km min}^{-1} $

### Question 5
# Explanation:
## Step1: Recall the formula for distance
The formula for distance $ d $ is $ d = v \times t $, where $ v $ is speed and $ t $ is time.
## Step2: Substitute the values
Given $ v = 65 \text{ km h}^{-1} $ and $ t = 3 \text{ h} $, so $ d = 65 \times 3 = 195 \text{ km} $
# Answer:
$ 195 \text{ km} $

### Question 6 (i)
# Explanation:
## Step1: Recall the formula for time
The formula for time $ t $ is $ t = \frac{d}{v} $, where $ d $ is distance and $ v $ is speed.
## Step2: Calculate time for first part
For first 30 km with speed 60 km h⁻¹: $ t_1 = \frac{30}{60} = 0.5 \text{ h} $
## Step3: Calculate time for second part
For next 30 km with speed 40 km h⁻¹: $ t_2 = \frac{30}{40} = 0.75 \text{ h} $
## Step4: Calculate total time
Total time $ t = t_1 + t_2 = 0.5 + 0.75 = 1.25 \text{ h} $ (or 75 minutes)
# Answer:
$ 1.25 \text{ h} $ (or 75 minutes)

### Question 6 (ii)
# Explanation:
## Step1: Recall the formula for average speed
Average speed $ v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} $
## Step2: Calculate total distance
Total distance $ d = 30 + 30 = 60 \text{ km} $
## Step3: Calculate average speed
From part (i), total time $ t = 1.25 \text{ h} $, so $ v_{avg} = \frac{60}{1.25} = 48 \text{ km h}^{-1} $
# Answer:
$ 48 \text{ km h}^{-1} $

### Question 7 (i)
# Explanation:
## Step1: Recall the formula for average speed
Average speed $ v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} $
## Step2: Calculate total distance
The train travels from A to B (200 km) and back from B to A (200 km), so total distance $ d = 200 + 200 = 400 \text{ km} $
## Step3: Calculate total time
Total time $ t = 2 + 3 = 5 \text{ h} $
## Step4: Calculate average speed
$ v_{avg} = \frac{400}{5} = 80 \text{ km h}^{-1} $
# Answer:
$ 80 \text{ km h}^{-1} $

### Question 7 (ii)
# Explanation:
## Step1: Recall the formula for average velocity
Average velocity $ \vec{v}_{avg} = \frac{\text{Displacement}}{\text{Total Time}} $
## Step2: Calculate displacement
Displacement is the shortest distance between initial and final position. Since the train starts from A and returns to A, displacement $ = 0 \text{ km} $
## Step3: Calculate average velocity
$ \vec{v}_{avg} = \frac{0}{5} = 0 \text{ km h}^{-1} $
# Answer:
$ 0 \text{ km h}^{-1} $

### Question 8 (i)
# Explanation:
## Step1: Recall the formula for speed
Speed $ v = \frac{\text{Distance}}{\text{Time}} $
## Step2: Convert distance to m
$ 1 \text{ km} = 1000 \text{ m} $, time $ t = 100 \text{ s} $
## Step3: Calculate speed
$ v = \frac{1000}{100} = 10 \text{ m s}^{-1} $
# Answer:
$ 10 \text{ m s}^{-1} $

### Question 8 (ii)
# Explanation:
## Step1: Recall the formula for velocity
Velocity $ \vec{v} = \frac{\text{Displacement}}{\text{Time}} $
## Step2: Determine displacement
Displacement is 1 km due east, which is 1000 m due east, time $ t = 100 \text{ s} $
## Step3: Calculate velocity
$ \vec{v} = \frac{1000 \text{ m (east)}}{100 \text{ s}} = 10 \text{ m s}^{-1} \text{ (east)} $
# Answer:
$ 10 \text{ m s}^{-1} $ (towards east)

### Question 9
# Explanation:
## Step1: Recall the formula for acceleration
Acceleration $ a = \frac{v - u}{t} $, where $ u $ is initial velocity, $ v $ is final velocity, and $ t $ is time.
## Step2: Substitute the values
Given $ u = 0 \text{ m s}^{-1} $ (starts from rest), $ v = 10 \text{ m s}^{-1} $, $ t = 2 \text{ s} $
## Step3: Calculate acceleration
$ a = \frac{10 - 0}{2} = 5 \text{ m s}^{-2} $
# Answer:
$ 5 \text{ m s}^{-2} $

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QUESTION IMAGE

Question

Explanation:

Question 3 (a)

Step1: Convert km to m and h to s

Step2: Calculate the value

Step1: Convert km to m and min to s

Step2: Calculate the value

Step1: Convert all speeds to \( \text{m s}^{-1} \)

Step2: Arrange in increasing order

Answer:

Question 3 (b)