 # Volume 4: Mechanics 1 Equations of Motion for Constant Acceleration.

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Volume 4: Mechanics 1 Equations of Motion for Constant Acceleration

We can use a velocity-time graph to find some equations that hold for a body moving in a straight line with constant acceleration. velocity (ms -1 ) time (s ) u v t0 Suppose when the time is 0... At any time, t, we let the velocity be v. the velocity is u. Ans: The gradient gives the acceleration. How do you find acceleration from a velocity-time graph. Constant acceleration means the graph is a straight line.

We can use a velocity-time graph to find some equations that hold for a body moving in a straight line with constant acceleration. velocity (ms -1 ) time (s ) u v t a  v  u t So, v  u 0 Suppose when the time is 0... the velocity is u. t From this equation we can find the value of any of the 4 quantities if we know the other 3. At any time, t, we let the velocity be v. Constant acceleration means the graph is a straight line.

a  v  u t a t  v  u v  u  a t Multiplying by t :  u  a t  v  We usually learn the formula with v as the “subject”. The velocity, u, when t  0, is often called the initial velocity.

e.g. A particle has an initial velocity of 40 m s -1 and moves in a straight line with a constant acceleration of 10 m s -2. Find the velocity after 5 seconds. Solution:

u = 40, e.g. A particle has an initial velocity of 40 m s -1 and moves in a straight line with a constant acceleration of 10 m s -2. Find the velocity after 5 seconds.

Solution: a = 10, e.g. A particle has an initial velocity of 40 m s -1 and moves in a straight line with a constant acceleration of 10 m s -2. Find the velocity after 5 seconds. u = 40,

Solution: t = 5 e.g. A particle has an initial velocity of 40 m s -1 and moves in a straight line with a constant acceleration of 10 m s -2. Find the velocity after 5 seconds. a = 10,u = 40,

Solution: t = 5 v  u  a t  v  40 + 10  5  v  90 m s -1 e.g. A particle has an initial velocity of 40 m s -1 and moves in a straight line with a constant acceleration of 10 m s -2. Find the velocity after 5 seconds. a = 10,u = 40, We can also use the velocity-time graph to find a formula which involves displacement.

velocity (ms -1 ) time (s ) u v t0 How do you find displacement from the graph. Ans: Displacement is given by the area under the graph. v  u  a t - - - - - - (1)

velocity (ms -1 ) time (s ) u v t 0 The usual letter for displacement is s. Using the area of the trapezium, v s t Substituting for v from equation (1) : v  u  a t - - - - - - (1) s s  (u  u  at )t 2 1 (2u  at )t  2 1  (2ut  at 2 ) 2 1 s  ut  at 2 2 1  (u  v) t 2 1 s s  - - - - (2) (u  v) t 2 1 s s 

v  u  a t - - - - - - (1) s  ut + at 2 - - - - - - (3) 2 1 So, by using the gradient and area from the velocity/time graph for constant acceleration, we have - - - - (2) (u  v) t 2 1 s s 

e.g.A particle moves in a straight line with constant acceleration. (a)If its initial velocity is 5 m s -1 and the acceleration is  0·5 m s -2, what is its displacement from the start after 20 s ? (b)On another occasion the particle starts from rest. How long does it take to reach a point 7 m from the start if the final velocity is 2 m s -1 ?

(a)If its initial velocity is 5 m s -1 and the acceleration is  0·5 m s -2, what is its displacement from the start after 20 s ? s, u, v, a, t Method: We have 4 equations to choose from so we list the letters for all 5 quantities and mark the ones we know and the one we want to find.

(a)If its initial velocity is 5 m s -1 and the acceleration is  0·5 m s -2, what is its displacement from the start after 20 s ? s, u, v, a, t Method: We have 4 equations to choose from so we list the letters for all 5 quantities and mark the ones we know and the one we want to find.

(a)If its initial velocity is 5 m s -1 and the acceleration is  0·5 m s -2, what is its displacement from the start after 20 s ? s, u, v, a, t Method: We have 4 equations to choose from so we list the letters for all 5 quantities and mark the ones we know and the one we want to find.

(a)If its initial velocity is 5 m s -1 and the acceleration is  0·5 m s -2, what is its displacement from the start after 20 s ? s, u, v, a, t ? Method: We have 4 equations to choose from so we list the letters for all 5 quantities and mark the ones we know and the one we want to find.

(a)If its initial velocity is 5 m s -1 and the acceleration is  0·5 m s -2, what is its displacement from the start after 20 s ? s, u, v, a, t ? Method: We have 4 equations to choose from so we list the letters for all 5 quantities and mark the ones we know and the one we want to find.

(a)If its initial velocity is 5 m s -1 and the acceleration is  0·5 m s -2, what is its displacement from the start after 20 s ? s, u, v, a, t ? We need the equation without v. v  u  a t s  ut + at 2 2 1 s  (u + v)t 2 1 Method: We have 4 equations to choose from so we list the letters for all 5 quantities and mark the ones we know and the one we want to find.

v  u  a t s  ut + at 2 2 1 s  (u + v)t 2 1 (a)If its initial velocity is 5 m s -1 and the acceleration is  0·5 m s -2, what is its displacement from the start after 20 s ? s, u, v, a, t ? We need the equation without v. s  ut + at 2 2 1  s  (5)(20) + (  0·5)(20) 2 2 1  100  100  0 After 20 s the displacement is zero. The particle is back at the start. Method: We have 4 equations to choose from so we list the letters for all 5 quantities and mark the ones we know and the one we want to find.

(b)On another occasion the particle starts from rest. How long does it take to reach a point 7 m from the start if the final velocity is 2 m s -1 ? Solution: s, u, v, a, t Tell your partner which letters give the quantities we know and the one we want to find. Ans...

(b)On another occasion the particle starts from rest. How long does it take to reach a point 7 m from the start if the final velocity is 2 m s -1 ? Solution: s, u, v, a, t

(b)On another occasion the particle starts from rest. How long does it take to reach a point 7 m from the start if the final velocity is 2 m s -1 ? Solution: s, u, v, a, t ?

(b)On another occasion the particle starts from rest. How long does it take to reach a point 7 m from the start if the final velocity is 2 m s -1 ? Solution: s, u, v, a, t ?

(b)On another occasion the particle starts from rest. How long does it take to reach a point 7 m from the start if the final velocity is 2 m s -1 ? Solution: s, u, v, a, t ?

(b)On another occasion the particle starts from rest. How long does it take to reach a point 7 m from the start if the final velocity is 2 m s -1 ? Solution: s, u, v, a, t ? We need the equation without a. v  u  a t s  ut + at 2 2 1 s  (u + v)t 2 1

(b)On another occasion the particle starts from rest. How long does it take to reach a point 7 m from the start if the final velocity is 2 m s -1 ? Solution: s, u, v, a, t ? We need the equation without a.  7  (0 + 2)t 2 1  t  7 s  (u + v)t 2 1 The particle takes 7 s. v  u  a t s  ut + at 2 2 1 s  (u + v)t 2 1

v  u  a t - - - - - - (1) s  ut + at 2 - - - - - - (3) 2 1 s  (u + v)t - - - - - - (2) 2 1 All the above contain t, so we now find an equation without t. From (1), v  u  a t  t  v  u a Substitute in (2), Multiply by 2a: 2as  (u  v)(v  u) 2as  uv  v 2  u 2  uv u 2  2as  v 2 Rearranging : v 2  u 2  2as  ------ (5)   v  u a 2 1 s  (u + v)

v  u  a t - - - - - - (1) v 2  u 2  2as  ------ (4) v  u  a t - - - - - - (1) s  ut + at 2 - - - - - - (2) 2 1 There are 4 equations that we can use to solve problems where the acceleration is constant. s  (u + v)t - - - - - - (3) 2 1 N.B.All the equations have come from  a displacement/time graph, so a graph or the equations can be used to solve a problem – although sometimes one method is easier than the other.

SUMMARY  The equations of motion for constant acceleration are There are only 2 different formulae here. The others can be found by combining (1) and (2) in different ways.  In constant acceleration problems the following letters are used: s : displacement t : time u : velocity when t  0 a : acceleration v : velocity at time t students usually value ancient teachers The formulae are not on the formulae sheets. v  u  a t - - - - - - (1) s  (u + v)t - - - - - - (2) 2 1 v 2  u 2  2as  - - - - - - (5) s  ut + at 2 - - - - - - (3) 2 1

s  ut  at 2 2 1 s  (u + v)t 2 1 v 2  u 2  2as A particle travels in a straight line with constant acceleration. Its initial velocity is 10 m s -1 and after 40 s its velocity is 15 m s -1. EXERCISE Find (a) the magnitude of the acceleration and (b) the displacement after 40 s. Solution: s, u, v, a, t ? v  u  a t 15  0  a(40)  5  40a  a  0·125 m s -2  (a) (b) We can use any of the 4 remaining equations: e.g. s  ut + at 2 2 1 s  10  40 + 0·125(40) 2  2 1 s  500 m 

The next part uses the section “2. Unit Vectors” from “Vectors in Mechanics”. Vectors in Mechanics Select from the options below. Continue

The equations of motion for constant acceleration can be adapted for use with unit vectors. The most common equations are: v  u  a t s  ut + at 2 2 1 Remember ! t is not a vector s  vt  at 2 2 1 s  (u + v)t 2 1 v 2  u 2  2as These equations can also be used with unit vectors: However, this is NOT a vector equation:

e.g. A particle is travelling with an acceleration given by a  (  2   7 ) m s -2 i j Find (a) the velocity and (b) the displacement of the particle at time t = 3 s. v  u  a t  v  3   ( 2  7 ) 3 i j i v  (  9   21 ) m s -1 ij v  3   6  21 i j i Solution: (a) Not all situations that you will meet will involve constant acceleration so check before you use the equations for constant acceleration. Initially the particle is at the origin with a velocity of 3 m s -1 i This is constant. An expression containing t would be variable.

e.g. A particle is travelling with an acceleration given by a  (  2   7 ) m s -2 i j Initially the particle is at the origin with a velocity of 3 m s -1 Find (a) the velocity and (b) the displacement of the particle at time t = 3 s. i Solution: s  (  18   31·5 ) m ij s  9   9  31·5 i j i (b) s  ut + at 2 2 1  s  3  (3   ( 2  7 ) 3 2 i j i 2 1

The summary page follows in a form suitable for photocopying.

 The equations of motion for constant acceleration are There are only 2 different formulae here. The others can be found by combining ( 1 ) and ( 2 ) in different ways.  In constant acceleration problems the following letters are used: s : displacement t : time u : velocity when t  0 a : acceleration v : velocity at time t students usually value ancient teachers The formulae are not on the formulae sheets. v  u  a t - - - - - - (1) s  (u + v)t - - - - - - (2) 2 1 v 2  u 2  2as  - - - - - - (5) s  ut + at 2 - - - - - - (3) 2 1 s  vt  at 2 - - - - - - (4) 2 1 Summary EQUATIONS OF MOTION FOR CONSTANT ACCELERATION TEACH A LEVEL MATHS – MECHANICS 1

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