Describing Motion Around UsClass 9 Science NCERT Solutions

Given: Distance from home to shop = 250 m

Calculation of Total Distance Travelled: The total path covered by the father is as follows:

Total distance travelled = 250 m + 250 m + 250 m + 250 m = 1000 m

Calculation of Displacement: Displacement is the shortest distance between the initial and final positions, including direction.

Since the initial and final positions are the same, the net change in position is zero. Displacement = 0 m

Yes, it is possible for the scooter to be accelerating even if its speedometer reading is constant.

In this case, the speedometer reading is constant, which means the speed (magnitude of velocity) is not changing. However, if the girl is turning or moving along a curved path (like a circular road), the direction of her velocity is continuously changing. Since the direction of velocity is changing, the velocity itself is changing, and therefore, the scooter is accelerating.

Example: An object in uniform circular motion has a constant speed but is always accelerating because its direction of motion is constantly changing. This acceleration is directed towards the center of the circle.

Given: Initial velocity, $u = 0 \text{ m s}^{-1}$ (since the car starts from rest) Final velocity, $v = 24 \text{ m s}^{-1}$ Time interval, $t = 6 \text{ s}$

1. Calculation of Average Acceleration: Formula: The first equation of motion is $v = u + at$. We can rearrange it to find acceleration: $a = \frac{v - u}{t}$.

Calculation: $a = \frac{24 \text{ m s}^{-1} - 0 \text{ m s}^{-1}}{6 \text{ s}}$ $a = \frac{24}{6} \text{ m s}^{-2}$ $a = 4 \text{ m s}^{-2}$

2. Calculation of Distance Travelled: Formula: The second equation of motion is $s = ut + \frac{1}{2}at^2$.

Calculation: $s = (0 \text{ m s}^{-1} \times 6 \text{ s}) + \frac{1}{2} \times (4 \text{ m s}^{-2}) \times (6 \text{ s})^2$ $s = 0 + 2 \times 36 \text{ m}$ $s = 72 \text{ m}$

Alternatively, using the third equation of motion, $v^2 = u^2 + 2as$: $(24)^2 = (0)^2 + 2 \times 4 \times s$ $576 = 8s$ $s = \frac{576}{8} = 72 \text{ m}$

Given: Initial velocity, $u = 28 \text{ m s}^{-1}$ Final velocity, $v = 0 \text{ m s}^{-1}$ (since the motorbike stops) Distance travelled, $s = 98 \text{ m}$

1. Calculation of Acceleration: Formula: We can use the third equation of motion, $v^2 = u^2 + 2as$, as it relates $u, v, a,$ and $s$.

Calculation: $(0 \text{ m s}^{-1})^2 = (28 \text{ m s}^{-1})^2 + 2 \times a \times 98 \text{ m}$ $0 = 784 + 196a$ $-196a = 784$ $a = -\frac{784}{196}$ $a = -4 \text{ m s}^{-2}$ The negative sign indicates that the acceleration is in the opposite direction to the velocity (deceleration).

2. Calculation of Time Taken: Formula: We can use the first equation of motion, $v = u + at$, to find the time.

Calculation: $0 \text{ m s}^{-1} = 28 \text{ m s}^{-1} + (-4 \text{ m s}^{-2}) \times t$ $0 = 28 - 4t$ $4t = 28$ $t = \frac{28}{4} \text{ s}$ $t = 7 \text{ s}$

Yes, objects A and B have equal velocity at one instant in time.

The correct options are (i) and (iii).

Given: Initial velocity, $u = 54 \text{ km h}^{-1}$ Final velocity, $v = 36 \text{ km h}^{-1}$ Time interval, $t = 36 \text{ s}$

Step 1: Convert velocities to m/s To convert km/h to m/s, we multiply by $\frac{1000}{3600} = \frac{5}{18}$. $u = 54 \times \frac{5}{18} \text{ m s}^{-1} = 3 \times 5 = 15 \text{ m s}^{-1}$ $v = 36 \times \frac{5}{18} \text{ m s}^{-1} = 2 \times 5 = 10 \text{ m s}^{-1}$

Step 2: Calculate the distance travelled To Find: Distance travelled, $s$

Since the acceleration is constant, we can use the equation that relates displacement, initial velocity, final velocity, and time.

Formula: $s = \frac{1}{2}(u + v)t$

Calculation: $s = \frac{1}{2}(15 \text{ m s}^{-1} + 10 \text{ m s}^{-1}) \times 36 \text{ s}$ $s = \frac{1}{2}(25) \times 36 \text{ m}$ $s = 25 \times 18 \text{ m}$ $s = 450 \text{ m}$

Final Answer: The distance travelled by the truck during this time is 450 m.

The motion of the car can be divided into three parts. We will calculate the distance travelled in each part and then add them to find the total distance.

Total Distance Travelled: Total distance, $S = s_1 + s_2 + s_3$ $S = 50 \text{ m} + 200 \text{ m} + 60 \text{ m} = 310 \text{ m}$

Final Answer: The total distance travelled by the car is 310 m.

To solve this problem, we need to calculate the total distance the bus travels from the moment the driver sees the obstacle until it comes to a complete stop. This total stopping distance is the sum of the distance travelled during the reaction time and the distance travelled while braking.

Step 1: Convert initial velocity to m/s Initial velocity, $u = 36 \text{ km h}^{-1}$ $u = 36 \times \frac{5}{18} \text{ m s}^{-1} = 2 \times 5 = 10 \text{ m s}^{-1}$

Step 2: Calculate the distance travelled during reaction time During the reaction time, the bus moves at a constant velocity.

Formula: Using the third equation of motion, $v^2 = u^2 + 2as$.

Calculation: $(0)^2 = (10)^2 + 2 \times (-2.5) \times s_{brake}$ $0 = 100 - 5 \times s_{brake}$ $5 \times s_{brake} = 100$ $s_{brake} = \frac{100}{5} = 20 \text{ m}$

Step 4: Calculate the total stopping distance Total stopping distance = Distance during reaction + Distance during braking $S_{total} = s_{react} + s_{brake} = 5 \text{ m} + 20 \text{ m} = 25 \text{ m}$

Since $25 \text{ m} < 30 \text{ m}$, the total stopping distance is less than the distance to the obstacle.

Final Answer: Yes, the bus will be able to stop before reaching the obstacle. It will stop 5 meters before the obstacle ($30 \text{ m} - 25 \text{ m} = 5 \text{ m}$).

The concepts of motion and rest are relative. An object's state of motion depends entirely on the frame of reference from which it is being observed.

Conclusion: An object kept on the Earth can be considered to be at rest for all practical purposes in our daily lives, where the Earth itself is the implied frame of reference. However, in an astronomical context where the Sun or the center of the galaxy is the frame of reference, the same object is in a state of complex motion. Therefore, whether an object is at rest or in motion is not an absolute property but is relative to the observer's frame of reference.

The distance run by the girl can be estimated by calculating the area under the velocity-time graph.

Estimation Method: Approximating as a Triangle We can approximate the entire area under the curve as a single triangle to get a good estimate of the distance.

Formula for Area of a Triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

Calculation: Since the velocity is in km/h and the time is in hours, the resulting area will be in kilometers (km). $\text{Distance} \approx \text{Area} = \frac{1}{2} \times (1 \text{ h}) \times (12 \text{ km/h})$ $\text{Distance} \approx 6 \text{ km}$

Final Answer: Based on the graph, the estimated distance the girl ran is approximately 6 km.

First, we will draw the velocity-time graph for the car's motion. Then, we will find the total displacement by calculating the area under the graph.

Velocity-Time Graph: The graph will consist of a horizontal line segment followed by an inclined line segment.

The total displacement is the sum of the area of the rectangle (Part 1) and the area of the trapezium (Part 2).

Final Answer: The displacement of the car on the state highway in the 2 min 6 s time interval is 774 m.

Step 2: Plot the velocity-time graphs Since both cars start from rest and have constant acceleration, their velocity-time graphs will be straight lines passing through the origin (0,0).

(The hint to plot five points is a way to ensure the straight line is drawn correctly, but the line is fully defined by the start and end points).

Step 3: Calculate displacement using the graph The displacement is the area under the velocity-time graph.

Displacement of Car A in 5 s: The area is a triangle with base = 5 s and height = 5 m s⁻¹. $s_A = \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \text{ s} \times 5 \text{ m s}^{-1} = 12.5 \text{ m}$

Displacement of Car B in 10 s: The area is a triangle with base = 10 s and height = 3 m s⁻¹. $s_B = \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \text{ s} \times 3 \text{ m s}^{-1} = 15 \text{ m}$