## Motion in a Straight Line

# Accelerated Motion

- Equations of motion are \tt s = ut+\frac{1}{2}at^{2}

v = u + at

v^{2 }− u^{2 }= 2as

\tt S_{n} = u + a\left(n-\frac{1}{2}\right) - Initial velocity of freely falling body is zero.
- Final velocity of freely falling body V = gt.
- Final velocity of freely falling body wr.to distance \tt v=\sqrt{2gh}
- Distance travelled in n
^{th}sec \tt S_{n}=0+g\left(n-\frac{1}{2}\right). - Ratio of distances travelled in 1s , 2s , 3s .... = 1 : 4 : 9 : .......n
^{2}. - Ratio of distances travelled in 1
^{st}2^{nd}3^{rd}= 1 : 3 : 5 .......(2n – 1). - Body falls from height ‘h’ time of descent = \tt \sqrt{\frac{2h}{g}}
- For a vertically projected body The maximum height reached \tt H=\frac{u^{2}}{2g}
- Time of ascent \tt Ta=\sqrt{\frac{2H}{g}}=\frac{u}{g}
- Time of descent \tt Td=\sqrt{\frac{2H}{g}}=\frac{u}{g}
- Time of flight \tt Tf=2\sqrt{\frac{2H}{g}}=\frac{2u}{g}
- If as friction in considered ta < td.
- Body always travels g/2 m during last second of upward motion.
- Equation of motion of body projected from top as tower is =\tt H= -u \ t\ +\frac{1}{2}gt^{2}
- Velocity with which it striker \tt V=\sqrt{u^{2}+2gh}
- If a body is projected vertically up with ‘u’ from tower it reaches ground with velocity ‘nu’ \tt H=\frac{u^{2}}{2g}\left(n^{2}-1\right)
- A particle projected vertically up from top of tower “t
_{1}” sec to reach the ground another particle thrown down with same velocity takes “t_{2}” sec velocity of projection \tt u=\frac{g}{2}\left(t_{1}-t_{2}\right). Height of tower \tt h=\frac{1}{2}g\ t_{1}t_{2}. - For oblique projection Horizontal position with time x = u cos θ t.
- For oblique projection vertical position with time \tt y=u\sin\theta\ t-\frac{1}{2}\ gt^{2}
- Equation of trajectory Y = \tt \tan\theta x-\frac{g}{2u^{2}\cos ^{2}\theta}\ x^{2}
- Time of ascent \tt Ta=\frac{u\sin\theta}{g}
- Time of flight \tt Tf=\frac{2u\sin\theta}{g}
- Maximum Height \tt H=\frac{u^{2}\sin^{2}\theta}{2g}
- Range (Horizontal) \tt R=\frac{u^{2}\sin 2\theta}{g}
- Range is maximum when θ = 45° R =\tt \frac{u^{2}}{g}
- Velocity of projectile at any instant t \tt V=\sqrt{\left(u\cos\theta\right)^2+\left(u\sin\theta-gt\right)^{2}}
- Angle made by projectile with horizontal \tt Tan \ \alpha=\left(\frac{u\sin\theta-gt}{u\cos\theta}\right)
- Vertical component of velocity at half of maximum height = \tt \sqrt{\left(u\cos\theta\right)^{2}+\left(\frac{u\sin\theta}{\sqrt{2}}\right)^2}
- If the particle passes two points situated at same height h at t
_{1}and t_{2}time Then time of flight T = t_{1}+ t_{2}& \tt h=\frac{1}{2}g\ t_{1}t_{2} - Angular momentum of particle at maximum height about point of projection \tt L=mu\ \cos\theta\times\frac{u^{2}\sin^{2}\theta}{2g}
- Torque acting on a particle at max height = \tt mg\times\frac{R}{2}.
- For complementary angles range is same.
- Let h
_{1}and h_{2}are heights “T_{1}” and “T_{2}” time of flights \tt h_{1}+h_{2}=\frac{u^{2}}{2g}\ R=4\sqrt{h_{1}h_{2}}\ R=\frac{1}{2}\ g\ T_{1}T_{2} - Time of flight of oblique projection from inclined plane \tt T=\frac{2u\sin\left(\alpha-\beta\right)}{g\cos \beta}
- Range \tt R=\frac{u^{2}}{g\cos^{2}\beta}\left[sin\left(2\alpha-\beta\right)-\sin\beta\right]
- Time of flight of projectile thrown down the inclined plane \tt T=\frac{2u\sin\left(\alpha+\beta\right)}{g\cos \beta}
- Range down the plane \tt R=\frac{2u^{2}\cos\alpha\ sin\left(\alpha+\beta\right)}{g\cos^{2}\beta}
- Horizontal position of horizontal projectile X = ut
- Vertical position of horizontal projectile \tt y=\frac{1}{2}gt^{2}.
- Equation of projector of horizontal projectile \tt y=\frac{gx^{2}}{2u^{2}}
- Time of descent of horizontal projection \tt t=\sqrt{\frac{2h}{g}}
- Range of horizontal projection \tt R=u\sqrt{\frac{2h}{g}}
- Velocity of horizontal projectile at any instant \tt v=\sqrt{u^{2}+g^{2}t^{2}}
- Angle made by horizontal projectile after same time Tan α= \tt \frac{gt}{u}
- Velocity on reaching the ground \tt v=\sqrt{u^{2}+2gh}
- From a certain height if two bodies are projected horizontally with velocities ‘u
_{1}’ and ‘u_{2}’ in opposite direction then time after which their velocity vectors one perpendicular \tt t=\sqrt{\frac{u_{1}u_{2}}{g}} - Distance between the two bodies when their velocities are perpendicular \tt d=\sqrt{\frac{u_{1}u_{2}}{g}}(u
_{1}+u_{2}) - Time after which their displacement vectors are perpendicular \tt 2\sqrt{\frac{u_{1}u_{2}}{g}}
- Distance between then when their displacement vectors are perpendicular \tt 2 \sqrt{\frac{u_{1}u_{2}}{g}}(u
_{1}+u_{2}) - If a body covers equal distances in equal time intervals the motion is said to be uniform.
- If a body covers unequal distances in equal time intervals the motion is said to be non uniform.
- Equation of trajectory of oblique projection from top of a tower. \tt H=-u\sin\theta\ T\ +\frac{1}{2}gT^{2}
- Range in the above case R = (u cos θ)T.
- At maximum height of oblique projective angle between velocity and acceleration is 90°.
- At maximum height of oblique projection from tower angle between velocity and acceleration is 90° .

### View the Topic in this video From 01:17 To 54:38

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1. Acceleration (a) = \tt \frac{Change\ in\ velocity\left(\Delta \nu\right)}{Time\ interval \left(\Delta t\right)}

2. Average acceleration : \tt \overrightarrow{a}_{a\nu}=\frac{\Delta\overrightarrow{\nu}}{\Delta t}=\frac{\overrightarrow{\nu}_{2}-\overrightarrow{\nu}_{1}}{\Delta t}

3. Instantaneous acceleration : \tt \overrightarrow{a}=\lim_{\Delta t \rightarrow 0}\frac{\Delta \overrightarrow{\nu}}{\Delta t}=\frac{d\overrightarrow{\nu}}{dt}

4. Equations of Uniformly Accelerated Motion:

(i) v = u + at (ii) \tt s=ut+\frac{1}{2}at^{2} (iii) v^{2} = u^{2} + 2as

5. Distance travelled in nth second \tt S_{n}=u+\frac{a}{2}\left(2n-1\right)

6. If a body moves with uniform acceleration and velocity changes from u to v in a time interval, then the velocity at the mid point of its path = \tt \frac{\sqrt{u^{2}+\nu^{2}}}{2}