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Displacement–time graphs. 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 ).
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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 accelerationcan 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!
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?
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 9.81ms-2downwards. 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.
Equations of projectile motion An object in freefall: • moves at a constant horizontal (x) velocity ax = 0 • moves at a constant vertical (y) acceleration. ay = g The following equations can therefore be applied. Can you see how they have been derived? constant x velocity x = vxt vy = uy + gt vy2 = uy2 + 2gy suvat equations for uy and vywith a = g uy + vy y = uyt + ½gt2 y = t 2
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 8.2ms-1 at an angle of 34° to the horizontal. Will the ball clear the net if it is 2.3m away and 95cm high at this point? What assumptions should you make to solve this problem? • no air resistance • no spin • initial height is zero.
Height of a projectile We need to calculate the value of y at x = 2.3m and determine whether or not it is greater than 0.95m. What are the relevant equations of motion? 8.2ms-1 x = vXt 0.95m 34° y = uyt + ½gt2 2.3m First, use the x equation to calculate t when x is 2.3. 2.3 = 8.2 × cos34° × t t = 0.34s
Height of a projectile So the ball reaches x = 2.3m when t = 0.34s. Now substitute this value of t into the y equation to find y, and determine whether or not it is greater than 0.95m. 8.2ms-1 0.95m 35° 2.3m y = uyt + ½gt2 y = ((8.2 × sin34°) × 0.34) + (½ × -9.81 × 0.342) y = 0.99 m So y is greater than 0.95 and the ball clears the net!
