A LEVEL MATHS: Kinematics
For particles travelling at constant velocity, what is the formula for displacement?
s=vt
Use calculus to derive the formula for s=ut +½at²
v = ds/dt Therefore s= ∫v dt Substitute v=u+ at which would equal ∫(u + at) dt = ut + ½at + C When S=0 and t=0 0=u(0) + ½a(0)² + C Therefore C=0 So the formula: s=ut +½at² is found
Since vectors cannot be squared, which suvat equation cannot be used when describing 2D motion using vectors?
v²=u²+2as
For projectiles: If an object is projected upwards with a speed of 30m/s at an angle of 25 degrees, what would the components of this look like?
y= 30sin25 x= 30cos25
What does the gradient on a velocity-time graph equate to?
Acceleration
What does the gradient of a displacement-time graph equate to?
Velocity
Find the maximum height which a ball reaches when it is projected at 25 degrees at 30m/s from a height of 1.5m?
(↑) S: ? U: 30sin25 V: 0 (because when it reaches max V=O) A: -9.8m/s² T V²=U²+2as 0=(30sin25)² +2(-9.8)s s=8.20.. Total height above ground= 8.20 + 1.5 = 9.70
A particle sets off from the origin at t=0 and moves in a straight line. At time t seconds, the velocity of the particle is v m/s, where v= 9t -2t². Find maximum velocity of the particle.
1) First differentiate the velocity and equate it to 0: 9-4t= 0 Therefore t= 2.25 2) Then differentiate it again to calculate whether or not this is a maximum value 3) Then When you find out that it is insert it back into the original equation and find the mazimum value of v: v= 9(2.25) -2(2.25²) =10.125
Car A travels along a straight road at a constant velocity of 30m/s, passing point R at time t=0. Exactly 2 seconds later, second car B, travelling at 25m/s moves in the same direction from point R. Car B accelerates at a constant 2m/s². Show that, when the two cars are level, t² - 9t -46= 0
Remember that when the velocity is constant the acceleration is 0!
A stone is thrown horizontally with speed 10m/s from a height of 2m above the horizontal ground. Find the time taken for the stone to hit the ground and the horizontal distance travelled before impact. Use g=9.8m/s²
Separate your answer into vertical and horizontal components
Derive the formula for v²= u²+2as
Substitute t= v-u/a into s=½(u+v)t Therefore you end up with v²= u²+2as
Derive the formula for s= vt - ½at²
Substitute u= v -at into s=½(v+u)t Therefore you end up with s=vt -½at²
Derive the formula for s=ut + ½at²
Substitute v= u+ at into s=½(u +v)t Therefore you end up with s= ut + ½at²
Derive the formula for s= 1/2(u+v)t from the velocity time graph
The area beneath the v/t graph represents the displacement, s. Here the area is a trapezium, so just use the formula for the area of a trapezium
What does a flat-line/ zero gradient, equate to on a displacement-time graph?
The particle is stationary
What does the beneath a velocity-time graph equate to?
The total distance travelled
When there is constant velocity, describe the acceleration
There is 0 m/s² because the speed is constant
From which type of graph do we derive the SUVAT equations? Draw the graph
Velocity-Time Graph
With regards to projectiles what are the two most important things to remember when applying SUVAT?
-Acceleration only acts in the vertically due to gravity (9.8m/s²), therefore horizontally acceleration = 0m/s² -Time is the same in both directions (therefore it connects the two directions)
Derive the formula v=u + at from the velocity time graph
1) Because it is a v/t graph the gradient corresponds to the acceleration,a. The graph is a straight line, crossing the y axis at u. Therefore you would use the y=mx+c equation for a line. This should give you the formula v=u + at
Whenever you see maximum/minimum, wha should be your first thoughts?
Differentiation
At t=0, a particle has position vector (6i + 8j)m, relative to a fixed origin. The particle is travelling at constant velocity (2i -6j) m/s. Find its position vector at t=4s.
First find its displacement using s=vt 4x(2i-6j) =8i -24j Then add to original displacement (6i +8j) + (8i -24j) =14i -16j)
A cricket ball is projected with a speed of 30m/s at an angle of 25 degrees above the horizontal. Assume the ground is horizontal and the ball is struck from a point 1.5m above the ground Find the length of time the ball was at least 5m above the ground
First find the displacement of at least 5m above ground: s=5-1.5= 3.5 Set up SUVAT (↑) S: 3.5 U:30sin25 V: A:-9.8m/s² T:? s=ut+ 0.5at² 3.5= (30sin25)t + 0.5(-9.8)t t=0.31421... t=2.27324 Therefore Length of time at least 5m above the ground is 2.27...-0.314...= 1.96s (3sf)
When a question asks you for the direction of a vector, what is it asking you for?
It is asking for the angle (tan⁻¹(o/a)
Find the horizontal distance travelled by a ball before it hits the ground when it is projected at 25 degrees at 30m/s from a height of 1.5m
We first want to find how long the ball's air time was: (↑) S: -1.5 (because this is total displacement) U: 30sin25 V: A: -9.8m/s² T:? s=ut + 0.5at² t= 2.70s (reject negative answer) (→) S:? U: 30cos25 V: A: 0 T: 2.70s s=ut+0.5at² s= 73.4m
Under what circumstances does one use the SUVAT equations?
When there is constant acceleration
Under what circumstances is calculus used in kinematics?
When we have variable acceleration
Use calculus to derive the formula for v=u+at
a = dv/dt Therefore v=∫a dt which = at + C When when v=u, then t=0 therefore a(0) +C u= C Therefore v=u + at
