NEET Physics Motion In A Straight Line Notes

Motion In A Straight Line

Distance: The actual length of the path travelled by the particle.

Displacement: Change in position in a particular direction (or the minimum distance between 2 points).

Speed = \(\frac{\text { distance }}{\text { time }}\)

Velocity = \(\frac{\text { displacement }}{\text { time }}\)

Avg. velocity = \(\frac{\text { total displacement }}{\text { total time }}\)

⇒ \(=\frac{x_f-x_i}{t_f-t_i}=\frac{\Delta x}{\Delta t}\)

Where, xf is the final position, xi is initial position, f is final time and h is the initial time.

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Instantaneous Velocity

Limiting value of average velocity is called instantaneous velocity.

⇒ \(\mathrm{v}=\lim _{\Delta \mathrm{t} \rightarrow 0} \frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}=\frac{\mathrm{dv}}{\mathrm{dt}}\)

The slope of the x – t graph is the average velocity.

The slope of the v – t graph is acceleration

Area under the v – t graph is displacement.

Kinematic equations of motion for a body moving with uniform acceleration

v= u +at

x= ut + ½ at²

v² = u² + 2ax

x= \(\left(\frac{\mathrm{u}+\mathrm{v}}{2}\right) \mathrm{t}\)

Sn = \(\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1)\)

a= v \(v \frac{d v}{d x}\)

Where, u = initial velocity, v = final velocity, a = acceleration, t = time, x Sn= distance travelled in nth= displacement and second

Calculation of distance travelled in n^th second:

Sn = Distance travelled in ‘n’ seconds – distance travelled in ‘n – 1’ seconds

Sn = Xn- Xn-1

= \(\mathrm{un}+\frac{1}{2} \mathrm{an}^2-\left[\mathrm{u}(\mathrm{n}-1)+\frac{1}{2} \mathrm{a}(\mathrm{n}-1)^2\right]\)

= \(\mathrm{un}+\frac{1}{2} a n^2-\left[\mathrm{un}-\mathrm{u}+\frac{1}{2} a\left(\mathrm{n}^2+1-2 \mathrm{n}\right)\right]\)

= \(\mathrm{un}+\frac{1}{2} \mathrm{an}^2-\left[\mathrm{un}-\mathrm{u}+\frac{1}{2} \mathrm{an}^2+\frac{\mathrm{a}}{2}-\mathrm{an}\right]\)

= \(\mathrm{un}+\frac{1}{2} \mathrm{an}^2-\mathrm{un}+\mathrm{u}-\frac{1}{2} \mathrm{an}^2-\frac{\mathrm{a}}{2}+\text { an }\)

= \(\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1)\)

Note:

x =  \(\mathrm{ut}+\frac{1}{2} \mathrm{at}^2\)

x= \((\mathrm{v}-\mathrm{at}) \mathrm{t}+\frac{1}{2} \mathrm{at}^2\)

x= \(\mathrm{vt}-\mathrm{at}^2+\frac{1}{2} \mathrm{at}^2\)

x= \(\mathrm{vt}-\frac{1}{2} a \mathrm{t}^2\)

[ v= u +at, u= v-at]

(This equation can be used to find displacement (x) if final velocity (v) is known).

If a particle travels half the distance with velocity v₁ and the remaining velocity with v₂, then the average velocity is given by

⇒ \(\bar{v}=\frac{2 v_1 v_2}{v_1+v_2}\)

If a particle travels with velocity with vx for first half time, and the remaining half time with v2, then the average velocity is given by

⇒ \(\overline{\mathrm{v}}=\frac{\mathrm{v}_1+\mathrm{v}_2}{2}\)

Basci Formulae Of Diffrrentiiation

Where ‘a’ is a constant and ‘u’ is a function of ‘x’

Basic Formulae of Integration:

Definite integration

⇒ \(\int_a^b f(x) d x=\left.g(x)\right|_a ^b=g(b)-g(a)\)

Illustration:

Suppose f(x) = x2. Determine the value of the definite integral from x = 1 to x = 2.

⇒ \(\int_1^2 x^2 d x=\left[\frac{x^3}{3}\right]_1^2=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}\)

Example:

Example:

The position of an object moving along x-axis is given by x = a + bt² where a = 8.5 m, b = 2.5 ms^-2 and t is measured in seconds. What is its velocity at t = 0s and t = 2s. What is the average velocity between t = 2s and t = 4s?

Solution: 

Equations of motion for constant acceleration using method of calculus

Consider a body starts with initial velocity ‘u’ (at t=0) and moves with a constant acceleration ‘a’ and attains a speed v in time ‘t’.w.k.t.,

Sign Convention

“Upward Direction Is Taken + Y–axis”

Free Fall:

Consider a body dropped from the top of the building at t = 0s (i.e ., u= 0 m/s^-2) and g= – 10 ms^-2)

The ratio of the distance travelled in 1 second, 2 seconds, 3 seconds and so on are in the ratio, x_1s: x_2s: x_3s= 1 : 4 : 9 := 1² : 2² : 3²:

The ratio of the distance travelled in 1^st second, 2^nd second, 3^rd second and so on are in the ratio, x_1n: x_2nd: x_3RD …………= 1:3:9:………(Galileo’s law of odd numbers)

x-t, v-t and a-t graphs in the case of free fall are given below

A car starts from rest and acquires a speed v with uniform acceleration a. Then it comes to stop with uniform retardation β.

Distance Travelled in time ‘t’

x= \(x=\frac{1}{2}\left(\frac{\alpha \beta}{\alpha+\beta}\right) t^2\)

If a body is dropped from a height ‘h

The time required for the body to reach the ground is

t= \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\)

The velocity acquired by the body on reaching the ground is given by

v=  \(\sqrt{2 \mathrm{gh}}\)

Note:

Time of ascent (ta) = Time of descent (t_d)

Time of flight T = t_a + t_d

T= \(\frac{\mathrm{u}}{\mathrm{g}}+\frac{\mathrm{u}}{\mathrm{g}}\)

T= \(\frac{2 \mathrm{u}}{\mathrm{g}}\)

The stopping distance of vehicles moving with constant retardation is given by

⇒ \(D_s=\frac{u^2}{2 a}\)

i.e., the stopping distance is directly proportional to the square of the initial speed.