8CconstantAccel


 * Kinematics with Constant Acceleration**

When the acceleration is constant, we have a set of rules we can use:

math \\ . \qquad v=u+at \\ .\\ . \qquad s=ut+\frac{1}{2}at^2 \\ .\\ . \qquad s=vt-\frac{1}{2}at^2 \\ .\\ . \qquad v^2=u^2+2as \\ .\\ . \qquad s=\frac{1}{2} \big( u+v \big) t math

Where: ... ... s = displacement ... ... u = initial velocity ... ... v = final velocity ... ... a = acceleration (constant) ... ... t = time

These rules are all derived by solving: math . \qquad a=\dfrac{dv}{dt} math and then combining in different ways.

Obtaining the four rules is not in the course but is well within the capability of a Specialist Maths student. Go here to see how it is done.

Example 1

A car accelerates from rest at 2.0m/s 2.

a) Find velocity after 15s We know: a = 2, u = 0, t = 15, v = ?

math \\ \text{Use } v=u+at \\ .\\ . \qquad v=0+2\times15=30 \; m/s math

b) Find distance travelled after 10s Has direction of travel (velocity) changed in that time? No!  So distance travelled equals displacement.  We know: a = 2, u = 0, t = 10, s = ?

math \\ \text{Use } s=ut+\frac{1}{2}at^2 \\. \\ . \qquad s=0\times10 + \frac{1}{2}\times 2 \times 10^2 \\. \\ . \qquad s =100m math

c) Find the displacement in the third second. **"In the third second"** means between t = 2 and t = 3  We know: a = 2, s(2) = ?, s(3 )= ?

math \\ \text{Use } s=ut+\frac{1}{2}at^2 \\. \\ . \qquad s(2)=0 \times 2+\frac{1}{2} \times 2 \times 2^2 =4m \\ .\\ . \qquad s(3)=0 \times 3+\frac{1}{2} \times 2 \times 3^2 =9m math

Displacement in 3rd second math \\ . \qquad = s(3) - s(2) \\. \\ . \qquad = 9 - 4 \\. \\ . \qquad = 5m math

Vertical Motion Under Gravity

If we ignore air resistance, gravity causes unsupported bodies to accelerate towards the ground with a constant acceleration of g = 9.8m/s 2. {Near the earth's surface}

Because the acceleration is constant, the rules listed above can be used.

Example 2

A ball is thrown from ground level vertically up with a velocity of 19.6m/s. {Ignore air resistance}

a) Find its maximum height. {maximum height occurs when v = 0}  We know: u = 19.6, v = 0, a = –9.8, s = ?

math \\ \text{Use } v^2=u^2+2as \\. \\ . \qquad 0^2=19.6^2 + 2 \times -9.8 \times s \\. \\ . \qquad 19.6s = 19.6^2 \\. \\ . \qquad s = 19.6m math

b) Find the time taken to reach the maximum height We know: u = 19.6, v = 0, a = –9.8, s = 19.6, t = ?

math \\ \text{Use } v=u+at \\. \\ . \qquad 0=19.6+-9.8 \times t \\. \\ . \qquad 9.8t=19.6 \\. \\ . \qquad t=2 \; \text{ seconds} math

c) Find how long the ball is in the air Reaches top at t = 2.  Symmetrical so takes another 2 seconds to fall

So hits ground at t = 4 seconds

math \text{Use } s = ut + \frac{1}{2}at^2 math
 * OR: **

We know: u = 19.6, a = –9.8, s = 0, t = ?

math \\ . \qquad 0 = 19.6t + \big( \frac{1}{2} \big) \big( -9.8 \big) \big( t^2 \big) \\. \\ . \qquad 0 = 19.6t - 4.9t^2 \\. \\ . \qquad 0 = t \big( 19.6 - 4.9t \big) \\. \\ . \qquad t = 0 \;\; \textit{ or } \;\; 19.6-4.9t = 0 math

t = 0 is when the ball was first thrown! so ball lands when:

math \\ . \qquad 19.6 - 4.9t = 0 \\. \\ . \qquad 4.9t = 19.6 \\. \\ . \qquad t = 4 \; \text{ seconds} math

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