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Position formula is the second equation in the equations of motion. It shows the relationship between position and time. There are mainly three equations of motion which take into account the velocity, displacement, acceleration and time during the motion of an object. These equations of motion can be derived using different methods like algebraic method, calculus method etc.
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Key Takeaways: Position formula, Displacement, Equations of Motion, Velocity, Acceleration, Speed
Position Formula
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We must be able to define the position x of any item in order to explain its motion. It reflects it at any given moment. To put it another way, we need to identify its position in relation to the standard frame of reference. The arbitrary set of axes from which the position and motion of an object are described is referred to as a frame of reference. When describing the position of an object in relation to stationary objects on Earth, we frequently use Earth as the frame of reference. We frequently employ reference frames that are not stationary but move around the Earth. As a result, when describing a person's position on a flight, we use the airplane as a reference rather than Earth.
Position Graph
Any object's true position is its exact coordinate or location as defined by its basic dimensions or other means. In other terms, Position refers to how far the location of your feature can deviate from its "True Position."
A rectilinear movement is one in which the trajectory is straight. Furthermore, this movement is carried out at a constant rate of acceleration. We put an origin s0 on the straight line, where an observer will measure the position s of the mobile at the time t. A polynomial function can be used to relate the mobile's position s to time t.
Position = initial position + initial velocity × time + 1/2 × acceleration × (time)2
As a result, when something moves from one position to another, it is referred to as displacement. Assuming the movement of a ball from position s1 to position s2.
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| Average Velocity Formula | Angular Velocity Formula |
| Orbital Velocity | Escape Velocity and Orbital Velocity |
Formula for the Position:
The position change Δs (position formula) is determined as,
Δs = s2 - s1,
Where; s1 = first position, s2 = second position,
Δs = change of displacement.
If the body's position varies throughout time t, the change in position at any point in time t is represented as x(t).
The equation is written as:
s = s0 + v0t + a×t2/2
We have:
s = final position (the position at the end of some event)
s0 = initial position (the position at the beginning of some event)
v0 = initial velocity (the velocity at the beginning of some event)
t = time (the duration of the event)
a = acceleration
The equation is only valid when,
- Acceleration is constant.
- Motion is constrained to a straight line.
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Derivation of Position Formula by Calculus Method
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Velocity is the rate of change of displacement. Mathematically, this is expressed as;
v = ds/dt
Rearranging the equation, we get
ds = vdt
Substituting the first equation of motion in the above equation, we get,
ds = (v0 + at) dt
or, ds = (v0dt + atdt)
On further simplification, the equation becomes:
\(\int_0^s\)ds = \(\int_0^t\)v0dt + \(\int_0^t\)atdt
= v0dt + \(\frac{1}{2}\)at2
Alternatively, it can be written as,
s – s0 = v0dt + \(\frac{1}{2}\)at2
or, s = s0 + v0t + \(\frac{1}{2}\)at2
This gives us the position-time equation for constant acceleration, also known as the second equation of motion.
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Things to Remember
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- The motion of an object can be understood as the change in position of an object with respect to its surroundings in a given interval of time.
- The position formula is the second equation of motion which describes the relationship between position and time.
- There are three equations of motion.
- Displacement is directly proportional to time and velocity.
- Position formula is calculated as initial position + initial velocity × time + 1/2 × acceleration × (time)2.
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Sample Questions
Ques. A person travels 30m distance. Calculate the position of the person at the end time 6s if the initial velocity of the person is 4m/s and angular acceleration is 3 m/s2. (3 marks)
Ans. Given;
Initial velocity, v0 = 4m/s
Initial position, x0 = 30m
Acceleration, α = 3m/s2
Time, t = 6s
The change in position of the person at time t is
\(x(t)=\frac{1}{2}\alpha t^2+v_0+X_0\)
Or, x(6) = 0.5 × 3 × (6)2 + 4 × 6 + 30
Or, x(6) = 54 + 24 + 30
Or, x(6)= 108 m.
Ques. A body with an initial velocity of 8 m/s begins to accelerate in t = 0 at a rate of 6 m/s2. What distance does it travel for the next 20 seconds from the instant it begins to accelerate? (2 marks)
Ans. To achieve the distance traveled, use the equation described above. We define the initial position x0 = 0 m, because we want to know the distance from that point, v0 = 8 m/s, t = 20s, and a = 6 m/s2.
\(x(t)=\frac{1}{2}\alpha t^2+v_0+X_0\)
x = (8 m/s) (20s) +(6 m/s2) (20 s)2/2
x = 160 m + 1200 m
x = 1360 m.
Ques. A train travels at a constant speed of 50 m/s and passes a signal in red. 60 meters after passing, the signal begins to slow down at a rate of 2 m/s2. At what distance from the signal does it stop completely in 10 seconds? (3 marks)
Ans. To achieve the distance travelled, use the equation described above. We have x0 = 60 m, v0 = 70 m/s, t = 20s, and a = -2 m/s2.
\(x(t)=\frac{1}{2}\alpha t^2+v_0+X_0\)
x = 60 m + (50 m/s) (10 s) + (-2 m/s2) (10 s)2/2
x = 60 m + 500 m - 100 m
x = 460 m.
Ques. A boy who has an initial velocity of 3ms-1, moves for a distance of 20 m. If the angular acceleration is 2 ms-2. Determine the position of the boy at the end of 5 sec. (5 marks)
Ans. Known parameters:
V0 (Initial velocity) = 3ms-1,
X0 (distance) = 20m,
a (angular acceleration) = 2ms-2,
t (time) = 5s
The alteration in the position of the boy at the instant of time t may be computed as:
X(t) = ½ at2 + v0t + X0
Thus,
X(6) = 0.5 × 252 + 3 × 5 + 20
= 25m + 15m + 20 m
= 60 m.
The position of the boy at the end of 5 sec will be 60 meters.
Ques. A boy who has an initial velocity of 2 m/s had already covered a distance of 10 m. If it has a constant acceleration of 2 m/s2, find the position of the boy at the end of 5s. (2 marks)
Ans. Position of the boy = x(t) = ½ at2 + v0t + X0
=½ x 2 x 25 + 2 x 5 + 10 = 45m.
Ques. Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 ms-1 and 30 ms-1. Verify that the graph shown in Fig. correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 ms-2. Give the equations for the linear and curved parts of the plot. (5 marks)
Ans. For first stone,
x (0) = 200 m, v (0) = 15 ms-1, a = -10 ms-2
x1 (t) = x (0) + v (0) t + 1/2 at2
x1 (t) = 200 + 15t – 5t2
When the first stone hits the ground, x1 (t) = 0
– 5t2 + 15 t+ 200 = 0
On simplification, t = 8 s
For second stone, x (0) = 200 m, v (0) = 30 ms-1, a = -10 ms-2
x1 (t) = 200 + 30t – 5t2
When this stone hits the ground, x1(t) = 0 .-. -5t2 + 30t + 200 = 0
Relative position of second stone w.r.t. first is given by x2 (t) – x1 (t) = 15t.
Since there is a linear relationship between x2(t) – x1 (t) and t, therefore the graph is a straight line.
For maximum separation, t = 8 s So maximum separation is 120 m
After 8 seconds, only the second stone would be in motion. So, the graph is in accordance with the quadratic equation.
Ques. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3ms-1 for 8s. How far does the boat travel during this time? (5 marks)
Ans. Since the motorboat starts from rest,
Its initial velocity will be 0.
Therefore,
Initial velocity= u = 0ms-1
acceleration= a = 3ms-2
time= t = 8s
Using the 2nd equation of motion to find distance,
s= ut + ½ at2
= 0 × 8 + ½ × 3 × (8)2
= 0 + ½ × 3 × 64
= 3 × 32
= 96 m.
AnswerQues. A racing car has a uniform acceleration of 4ms-2. What distance will it cover in 10s after start? (3 marks)
Ans. Since the car was initially at rest,
Initial velocity = u = 0ms-1
Acceleration = a = 4ms-2
Time = t = 10s
Finding distance covered using the 2nd equation of motion,
s= ut + ½ at2
= 0 × 10 + ½ × 4 × (10)2
= 0 + 2 × 100
= 200m.
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