Graphical Representation of Motion

Graphical representation of motion is representation of motion using graphs that show the relationship between various parameters such as time, displacement, velocity, and acceleration. Here are some common types of graphs used to represent motion.

Distance-Time graph:

• Distance-Time graphs show the change in position of an object for time.
• Linear variation = non-linear variations and uniform motion imply non-uniform motion
• The slope gives us speed

• As the slope is constant OA implies uniform motion with constant speed.
• AB implies the body is at rest as the slope is zero
• B to C is non-uniform motion

Velocity-Time Graph:

• Velocity-Time graphs show the change in velocity concerning time.
• Slope gives acceleration
• The displacement is given by the area below the curve.
• Line parallel to x-axis implies constant velocity.

• OA = constant acceleration, AB = constant velocity, BC = constant retardation

Also Read,

• Acceleration-Time Graphs
• Speed, Time and Distance

Equations of Motion

The motion of an entity moving at uniform acceleration can be described with the assist of three equations, namely

(i) v = u + at

(ii)[Tex] v^{2} – u^{2}= 2as[/Tex]

(iii) [Tex]s = ut + \frac{1}{2} at^{2}[/Tex]

Uniform Circular Motion

• It is called uniform circular motion if an object moves in a circular path with uniform speed.
• Velocity is varying as direction keeps altering.
• Acceleration is constant.

Sample Questions

Question 1. Two cars F and M race each other. The Car F drove for 2 minutes at a speed of 7.2 km/h, rested for 56 minutes and again ran for 2 minutes at a speed of 7.2 km/h. Find the average speed of the car F in the race.

Answer:

We know that

Distance= speed×time

Distance travelled in first 2 minute [Tex]= \frac{7.2\times 2}{60}= 0.24 km[/Tex]

Total distance = 0.24 + 0.24 = 0.48 km

Total time = 2+2+56 = 60 minute = 1 hr

Average speed = [Tex]\frac{0.48}{1}  [/Tex]=0.48 km/h

Question 2. A train 200 m long is moving with a velocity of 72 km/hr. Find the time it takes to cross the bridge which is 1 km long?

Answer:

Given Length of Train = 200m, velocity = 72 km/hr = 20 m/s, Length of the Bridge =1 Km.

Total distance covered by the train to fully pass the bridge = 1000 + 200 = 1200 m

So, time taken

Time[Tex]= \frac{distance}{velocity}= \frac{1200}{20}= 60 sec[/Tex]

Question 3. An object travels 26 m in 4 s and then another 22m in 3s. What is the average speed of the object?

Answer:

Total distance travelled by the object =26 m + 22 m = 48 m

Total time taken = 4 s + 3 s = 7 s

Average Speed = total distance travelled/total time taken

Average Speed[Tex]=\frac{48}{7}=  [/Tex] 6m/s

Question 4. Stopping distance of vehicles: When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity (v_{0}) and the braking capacity, or deceleration, that is caused by the braking. A car travelling at a speed of 72km/hr suddenly applies the brake with a deceleration of 10m/s². Find the stopping distance of the car.

Answer: 

Here u=72 km/hr [Tex]= \frac{72\times 1000}{3600}  [/Tex] m/s = 20 m/s, v=0

a= -10 m/s²

Now using the relation

[Tex]v^{2}= u^{2} + 2as[/Tex]

[Tex]0= 20^{2} + 2\times (-10)\times s[/Tex]

s = 20m

Question 5. A train starts from rest and accelerate uniformly at the rate of 10 m/s2 for 5 sec. Calculate the velocity of train in 5 sec.

Answer:

Here u=0, a= 10m/s², t= 5 sec, v=?

Now v= u + at

v = 0+10×5= 50 m/s

Question 6. An electron moving with a velocity of 5 × 10³ m/s enters into a consistent field and acquires a consistent acceleration of 2 ×10³ m/s² within the direction of the initial velocity.

i. Determine the time during which electron velocity is going to be doubled?

ii. How much distance electron would cover at this time?

Answer:

Given u = [Tex]5× 10^{3} m/s, a = 2×10^{3}   [/Tex] m/s²

[Tex]v = 2u = 10^{4}   [/Tex] m/s

i. Using v = u + at

[Tex]10^{4} = 5× 10^{3} + 2×10^{3}\times t[/Tex]

or, t = 2.5 sec

ii. Using[Tex] s = ut + \frac{1}{2} at^{2}[/Tex]

[Tex]s = 5× 10^{3} × 2.5 + \frac{1}{2} × 2×10^{3} × 2.5^{2}[/Tex]

[Tex] = 15.625\times 10^{3} m[/Tex]

We use general things around us that are moving, like if we see around us, monitor air moving around us, like we have clocks with the hands moving, we all know that day and night is caused because of motion of Earth around the Sun, yet seasons are caused because of it. So we are going to study in detail about what exactly the term motion is?

To recognize more about motion or rest we should first get known with the term reference point or stationary object. To comment on the state of any object, we have to consider a reference point or stationary object or surroundings. In this reference point, the stationary object does not change its position.

For Example: If we consider a pole and a car, then a pole will be stationary as it can’t change its location, but the car can alter. So, to say that car is at rest or motion we need to consider its motion concerning that of the pole or any other stationary object. This stationary object can also be called a reference point.

Rest and Motion

Let us consider a car standing in front of the house. Let’s say primarily it is at location A then after some time, it moves to location B. That means it has changed its position concerning the stationary object that is the house.

And if that car keeps on standing at position A that means it hasn’t changed its position concerning the house that means it is at rest so here we define the terms rest and motion.

Rest: When a body doesn’t change its position for its surroundings or reference point, then the body is said to be at rest.

Motion: When the body changes its position to its surroundings or reference point, then the body is in motion.

So, we can say that when an object is moving or in motion, it possesses the following characteristics as given below.

Characteristics of a moving object:

The moving object changes its position with time. Now we have seen that movement of the car can be easily seen that is we don’t have to concentrate much but to know the movement of an arm of clock let’s say hour hand we need to keep verify this is because some motion is that quick that we can see it happening, on the other hand, some motion is so slow that it can’t be seen undoubtedly.

We can elaborate it as: Our wristwatch has three hands: minute, seconds and hour hand .out of them second-hand moves fast that its motion can be seen but to see the motion in hour hand and minute hand we have to keep the track as the motion in them is moderately slow.