**7.1 Position :** a vector quantity – distance from a fixed origin

A vector is a quantity which has both magnitude and direction.

Displacement : a vector quantity -measured from the starting quantity (50 m West of starting point)

Distance travelled : a scalar quantity with no direction (125 m)

A scalar is just a number, it has no direction e.g time, mass, length

Average Velocity = \(\frac{Displacement}{Time Taken}\)

Speed = \(\frac{Total Distance}{Total Time}\)

Acceleration = \(\frac{Change of Velocity}{Time}\)

Distance and speed are always positive .

**7.2 Graphs in Kinematics**

**(a) Motion in a straight line**

**(b) Displacement v/s time**

**(c) Velocity v/s time**

**(d) Gradient under the graph**

**(e) Area under the graph**

** Example1: **A particle is initially travelling at a speed of 2 ms

Find the maximum speed and the time spent decelerating.

Sketch a speed-time graph.

If the total distance travelled is 1130 metres, find the time spent travelling at a constant speed.

*Solution: *

For maximum speed: *u = *2, *a* = 3, *t* = 10, v = u +at ⇒ c is maximum speed.

For deceleration from 32ms^{-1} at 2ms^{-1},the time taken is 32 ÷ 2 = 16 s.

Distance travelled in first 10 secs is area of trapezium = ½(2 + 32) × 10 =170 metres,

distance travelled in last 16 secs is area of triangle = ½ × 16 × 32 = 256 metres,

⇒ distance travelled at constant speed = 1130 – (170 + 256) = 704 metres

⇒ time taken at speed of 32 ms^(-1) is 704 ÷ 32 = 22 s.

**7.3 Constant acceleration in a straight line**

The equations of motion are :

\({v}\) = \({u + at}\)

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

\({v^2}\) = \({u^2 + 2as}\)

\({s}\) = \(ut+\frac{1}{2}(u +v)t\)

** Example2 : **A particle moves through a point O with speed 13ms

**Solution:**

u = 13, a = -6, s = 12, t = ?.

Use s = ut + ½at^{2} ⇒ 12 = 13t + ½ × (-6) × t^2

⇒ 3t^{2} – 13t + 12 = 0 ⇒ (3t – 4)(t – 3) = 0

⇒ t = \(ut+\frac{4}{3}at^2\) or 3.

Answer : Particle is 12 m from O after \(ut+\frac{4}{3}at^2\)or 3 s.

**Motion in 2D using vectors**

Addition of 2 vectors

If the vectors are \(\vec{u}\) = \(\dbinom{a}{b}\) and \(\vec{v} +\vec{v}\) = \(\dbinom{a + c}{b + d}\)

\(\vec{r}\) = \({3t}\hat{r} +8t\hat{j}\) (motion in 2D)

**7.4 Use of calculus for motion in a straight line**

r = f(t) \({v}\) = \(\frac{dr}{dt}\) \({a}\) = \(\frac{dv}{dt}\) = \(\frac{d^2r}{dt^2}\)

\({r}\) = \(\int{vdt}\) \({v}\) = \(\int{adt}\) a = f(t)

Motion in 2D

\(\hat{r}\) = \(t^3\hat{j} + t^{\frac{5}{3}}\hat{j}\)

**7.5 Motion under gravity**

With vectors

Projectiles

Initial velocity : \(\vec{u}\) = \(\vec{u}cosθ\hat{i} + \vec{u}sinθ\hat{j}\)

Acceleration : \(\vec{a}\) = \(-g\vec{j}\)

Velocity after t s : \(\vec{v}\) = \(\vec{u}cosθ\hat{i} +(\vec{u}sinθ -gt)\hat{j}\)

Particle moving in a horizontal direction reaches maximum height when \(\hat{j}\) = 0 => usinθ – t=0

Displacement after t s

\(\hat{r}\) = \(\vec{u}tcosθ\hat{i} +( \vec{u}sinθ\hat{j}\frac{g}{2}t^2)\hat{j}\)

If the projectile is launched from the ground, it will return to ground when \(\vec{r}\) = 0 => \(\vec{u}tsinθ -\frac{g}{2}t^2\) = 0

Range = total distance travelled in the horizontal direction \({R}\) = \(\frac{u^2sinθ}{g}\)

** Example3 :** A ball is thrown vertically upwards from

(a) Find the greatest height reached.

(b) Find the total time before the ball returns to *O*.

(c) Find the velocity after 2 seconds.

** Solution:** Take upwards as the positive direction.

(a) At the greatest height, *h*, the velocity will be 0 and so we have

*u* = 14, *v* = 0, *a* = –9.8 and *s* = *h* (the greatest height).

Using V^{2 }= u^{2 } + 2as we have 0^{2} –14^{2} = 2 × (–9.8) × h

⇒

h = 196 ÷ 19.6 = 10.

Answer: Greatest height is 10 m.

(b) When the particle returns to O the distance, s, from O is 0 so we have

s = 0, a = –9.8, u = 14 and t = ?.

Using s = ut + ½at^{2 } we have 0 = 14t – ½ × 9.8t^{2 }

⇒ t(14 – 4.9t) = 0

⇒ t = 0 (at start) or t = 26/7 s.

Answer: The ball takes 26/7 seconds to return to O.

(c) After 2 seconds, u = 14, a = –9.8, t = 2 and v = ?.

Using v = u + at we have v = 14 – 9.8 × 2

⇒ v = –5.6.

Answer: After 2 seconds the ball is travelling at 5.6 m s^{-1} downwards.