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Volume 4: Mechanics 1 Equations of Motion for Constant Acceleration

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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.

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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.

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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.

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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:

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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.

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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,

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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,

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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.

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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)

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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

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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

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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 ?

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(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.

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(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.

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(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.

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(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.

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(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.

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(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.

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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.

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(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...

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(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

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(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 ?

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(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 ?

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(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 ?

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(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

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(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

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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)

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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.

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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

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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

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

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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:

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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.

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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

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The summary page follows in a form suitable for photocopying.

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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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