Displacement–time graphs

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Acceleration – a change in velocity

Velocity changes when there is a change in its magnitude (i.e. a change in speed), a change in its direction, or both.So acceleration can include:

speeding up

slowing down (deceleration)

changing direction (e.g. centripetal acceleration)

So even though a geostationary satellite is travelling in a circle at a steady speed, it is actually accelerating as it constantly changes direction!


Velocity–time graphs


Four ‘suvat’ equations

Motion under constant acceleration can be described using the following four equations:

1. v = u + at

2. s = ut + ½at2

3. v2 = u2 + 2as

4. s = ½(u + v)t

These are known as the ‘suvat’ or constant acceleration equations, where u is the initial velocity, a is the acceleration, and s and v are the displacement and velocity at time t. How can these equations be derived?


Using the suvat equations


Analysing a velocity–time graph


Acceleration under gravity


Acceleration of freefall

An object that falls to the ground with no forces acting on it except gravity is said to be in freefall.

This can only occur when the effects of air resistance are negligible.

Any object in freefall, close to the Earth’s surface, experiences vertical acceleration of 9.81 ms-2 downwards. This is often denoted by the letter g.

‘Freefall’ includes both ‘rising’ and ‘falling’ motion, whether a projectile follows a parabola or a simple vertical line.


Capturing projectile motion


Equations of projectile motion

An object in freefall:

moves at a constant horizontal (x) velocity

moves at a constant vertical (y) acceleration.

ax = 0

ay = g

The following equations can therefore be applied. Can you see how they have been derived?

x = vxt

y =

vy = uy + gt

y = uyt + ½gt2

vy2 = uy

2 + 2gy

uy + vyt

constant x velocity

suvat equations for uy and vy with a = g

2


Birdman rally


Height of a projectile

A tennis player hits a volley just above ground level, in a direction perpendicular to the net.

The ball leaves her racquet at 8.2 ms-1 at an angle of 34° to the horizontal.

Will the ball clear the net if it is 2.3 m away and 95 cm high at this point?

What assumptions should you make to solve this problem?

no air resistance

no spin

initial height is zero.



We need to calculate the value of y at x = 2.3 m and determine whether or not it is greater than 0.95 m.

What are the relevant equations of motion?

x = vXt

y = uyt + ½gt2

First, use the x equation to calculate t when x is 2.3.

2.3 = 8.2 × cos34° × t t = 0.34 s

34°

8.2 ms-1

0.95 m

2.3 m



Now substitute this value of t into the y equation to find y, and determine whether or not it is greater than 0.95 m.

y = ((8.2 × sin34°) × 0.34) + (½ × -9.81 × 0.342)

So y is greater than 0.95 and the ball clears the net!

35°

8.2 ms-1

0.95 m

2.3 m

So the ball reaches x = 2.3 m when t = 0.34 s.

y = uyt + ½gt2

y = 0.99 m


Interactive cannon


Glossary


What’s the keyword?


Multiple–choice quiz

Displacement–time graphs

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