Chapter 2~ Accelerated Motion

Question 1

Definition of acceleration

Answer 1

°the rate of change in velocity

Question 2

Formula for acceleration =

Answer 2

a=🔼v/ 🔼t

Question 3

🔼v stands for

Answer 3

The change in velocity.

🔼v = v (final) - u(initial)

Question 4

Any object has acceleration if (2)

Answer 4

• it’s speed is changing

* changing its direction

Question 5

Unit of acceleration

Answer 5

ms-2

Question 6

Gradient of a velocity time ⌚ graph =

Answer 6

Acceleration

Question 7

A straight line with a positive slope on a velocity time ⌚ graph represents

Answer 7

Constant acceleration

Question 8

The steeper the slope of a v-t graph…

Answer 8

The greater the acceleration

Question 9

When the gradient on a v-t graph is zero. It means two things

Answer 9

1. velocity is constant

2. Therefore acceleration =0

Question 10

A negative slope on a v-t graph means

Answer 10

Deceleration

Question 11

If the slope is changing on a v-t graph it means

Answer 11

Acceleration is changing

Question 12

The area under a v-t graph equals?

➕ formula

Answer 12

Displacement

Area =1/2 x b x h

Sometime is a rectangle A= L x B

(sometimes you have to count the ‘squares’ under the graph)

Question 13

The 4 equations of motion:

Answer 13

1. v= u + at
2. v^2= u^2 +2as
3. s= ut + 1/2at^2
4. s= (u+v/2) x t

Question 14

The 4 equations of motion can only be used for 2 things namely:

Answer 14

1. Motion in a straight line

2. For an object with constant acceleration

Question 15

The 4 equations of motion can only be used for 2 things namely:

Answer 15

1. Motion in a straight line

2. For an object with constant acceleration

Question 16

How to derive the first equation of motion:

v= u + at

2 steps :

Answer 16

1. Acceleration is defined as the rate of change in velocity :

a= v-u/t

2. Rearrange this to get v as the subject of the formula :
   v =u +at

Question 17

How to derive s=((u+v) /2)x t

2 steps

Answer 17

1. Displacement =the average velocity x time taken (defined from v =d/t)
2. Average velocity = (v+u) /2

Therefore :

s= ((u+v) /2) x t

Question 18

How to derive s = ut +1/2at^2

3 steps :

Answer 18

Step 1: We need 2 equations :

1. v=u+at
2. s= ((u+v) /2)x t

Step 2: Sub v from equation 1 in equations 2’s v

s= ((u+u+at) /2)x t

s=(2ut)/2 +(at^2)/2

Step 3: simplify

So s= ut + 1/2 at^2

Question 19

How to derive v^2 =u^2 + 2as

3 steps :

Answer 19

1. Use two equations :

One : v= u+at

Two : s= ((u+v)/2)x t

2. Sub t in from equation one

s= ((u +v) /2) +((u-v) /a)

Step 3

Rearrange to get 2as =(u+v) (v-u)

Then v^2 = u^2 +2as

Question 20

For s=ut +1 /2at^2

What 2 equations do you use and what do you substitute?

Answer 20

Equations: v =u +at

                s= ((u+v) /2) x 2

Sub v in from equation one in equation two

Question 21

v^2 = u ^2 +2as

What 2 equations do you use and what do you substitute in?

Answer 21

Equations : v = u +at

                   s = ((u+v) /2) x t

Sub in t from equation one into equation two

Question 22

What is acceleration caused by gravity

Answer 22

9.81ms-^2

Question 23

The vertical motion of an object is affected by 1 things :

Answer 23

1. Weight

Question 24

The horizontal component is unaffected by?

Answer 24

Gravity

Question 25

In the absence of air resistance, the object has a constant velocity in the horizontal when means….

Answer 25

No acceleration

Question 26

To find the component of any vector (e.g. Displacement, velocity, acceleration, etc) in a particular direction, there are 2 steps :

And the equation

Answer 26

Step 1:
find the angle🎈 between the vector and the direction of interest

Step 2 :
Multiple the vector by cosine of the angle🎈

vcos🎈

Question 27

Definition of terminal velocity

Answer 27

The maximum velocity of an object travelling through a fluid. The resultant force on the object is zero. (when the weight of the object = the air resistance)

Question 28

Definition of lost volts

Answer 28

The difference between the e.m.f. and the terminal p.d in a circuit. It is equal to the voltage ⚡ across the internal resistance.