Flashcards · 13 cards

Kinematics & dynamics

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The four equations for constant acceleration?
Answer
$v = u + at$ , $s = \frac12(u + v)t$ , $s = ut + \frac12at^2$ and $v^2 = u^2 + 2as$ .
From the lesson: Equations of motion and velocity–time graphs
Rule
On a velocity–time graph, what do the gradient and the area give?
Answer
The gradient is the acceleration; the area under the graph is the distance travelled.
Constant accelerationgradient = a = (v − u) ÷ t; area = s = ½(u + v)t
From the lesson: Equations of motion and velocity–time graphs
Rule
The average speed for a journey in several stages?
Answer
Total distance (the area under the graph) ÷ total time.
A journey in three stagesdistance = area; average speed = distance ÷ time
From the lesson: Equations of motion and velocity–time graphs
Rule
A ball is thrown straight up at $u$ . Its greatest height, and the time to reach it?
Answer
At the top $v = 0$ : greatest height $\dfrac{u^2}{2g}$ , reached after $\dfrac ug$ . It takes as long again to come down.
Thrown straight upAt the top v = 0; greatest height = u² ÷ 2g
From the lesson: Motion under gravity and motion from a formula
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The acceleration due to gravity in these questions?
Answer
$g = 10$ m s⁻² unless told otherwise, always acting downwards.
From the lesson: Motion under gravity and motion from a formula
Rule
Displacement is given as a formula in $t$ . How do you get velocity and acceleration?
Answer
Differentiate: $v = \dfrac{ds}{dt}$ , $a = \dfrac{dv}{dt}$ . Going back, $v = \int a\,dt$ and $s = \int v\,dt$ .
Displacement, velocity, accelerationv = ds/dt, a = dv/dt; v = ∫a dt, s = ∫v dt
From the lesson: Motion under gravity and motion from a formula
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Newton's second law?
Answer
$F = ma$ , where $F$ is the resultant force (newtons = kg × m s⁻²).
From the lesson: Force and acceleration
Rule
In a lift accelerating upwards at $a$ , the reaction on a person of mass $m$ ?
Answer
$R - mg = ma$ , so $R = m(g + a)$ .
In a lift accelerating upwardsR − mg = ma, so R = m(g + a)
From the lesson: Force and acceleration
Rule
A block is pushed with force $P$ along a rough floor. The equation of motion?
Answer
$P - \mu R = ma$ , with $R = mg$ .
Pushed along a rough floorP − μR = ma, with R = mg
From the lesson: Force and acceleration
Rule
Two masses hang over a smooth pulley. The equations, and the acceleration?
Answer
$m_1g - T = m_1a$ and $T - m_2g = m_2a$ . Adding them: $a = \dfrac{(m_1 - m_2)g}{m_1 + m_2}$ .
Two masses over a pulleym₁g − T = m₁a and T − m₂g = m₂a
From the lesson: Force and acceleration
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Writing equations of motion with masses: what must you remember?
Answer
Change each mass to a weight, $mg$ : a 6 kg mass pulls with 60 N. And the tension equals the hanging weight only when nothing accelerates.
From the lesson: Force and acceleration
Which method?
WAEC 2013 · Paper 2 · Q8
The diagram shows a light inextensible string that passes over a smooth pulley and carries masses of $6\text{ kg}$ and $5\text{ kg}$ at its ends. When the system is released from rest: $[\text{Take } g = 10\text{ m s}^{-2}]$
Not to scale.
find the acceleration of each mass;
calculate, correct to two decimal places, the distance moved in 2 seconds.
Which way does each mass move, and what are the equations?
Answer
The heavier 6 kg mass goes down. $60 - T = 6a$ for it and $T - 50 = 5a$ for the other; add them to remove $T$ .
From the lesson: Force and acceleration Try the question
Which method?
WAEC 2016 · Paper 2 · Q15
A particle is projected vertically upwards from the ground with speed $30\text{ m s}^{-1}$ . Calculate the: (i) maximum height reached by the particle; (ii) time taken by the particle to return to the ground; (iii) time(s) taken for the particle to attain a height of $40\text{ m}$ above the ground. $[\text{Take } g = 10\text{ m s}^{-2}]$
What do you use at the greatest height?
Answer
The velocity there is 0: $0 = 30^2 - 2(10)s$ . The time up is $\frac{30}{10}$ s, and it takes as long to come down.
From the lesson: Motion under gravity and motion from a formula Try the question