
[Ended]Physics Tutoring Session #10 Motion Questions (Continuation) on Thurs. 3rd March., 2016 @ 4:30  5:30pm
Physics
Chima Sunday (Tutor)
03032016 15:17:00 +0000This session has ended for today
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Survey Feedback Link  http://goo.gl/forms/Tasr40Ck1ZParameters of Motion
i) Displacement is defined as the distance travelled in a specified direction
ii)Speed v is the rate of change of distance with time. v= s/t. Speed is said to be uniform if s/t is constant through out the journey.
Velocity is speed calculated in a given direction[recall that distance is for speed and rate of change of displacement is for velocity]
iii)Acceleration a is the rate of change of velocity increase with time While De acceleration is the rate of change of velocity decrease with time. a = v/t.
A motion is said to be a uniform acceleration if v/t is constant through out the journey e.g is the acceleration due to gravity.Equations of Motion
If a body starts with initial velocity u, accelerates uniformly along a straight line with acceleration a and covers a distance s in a time t when it velocity reaches a final velocity v. then
s = (v+u/2)t ...............................(1)
a = vu/t; v = u + at ......................(2)
Eliminating t in (1) and (2) to obtain
v^{2} = u^{2} + 2as ........................... (3)
Eliminating v in (1) and (2) to have
s = ut + (1/2)at^{2}
.................................(4) These four equations are used in solving problems associated with uniformly accelerated motion.Projectile Motion
Soved Examples
Projectile motion is a type of motion that carries out two independent motions (a) a constant horizontal motion (b) a vertically downward acceleration of free fall due to gravity. E.g of such motion is a stone shot out from a catapult, kicked football, thrown javelin, high jumper etc.
Parameters We use in Defining Projectile Motion
1) Time of flight (T).
This is the time required for for a projected body to return to the same level from which it was projected.
To calculate this parameter, we need the initial velocity u and angle of projection θ (usually measured from the horizontal)
For we to find this, we need to resolve the initial velocity to its vertical component U_{y}
and from our knowledge of resolution of vectors u_{y} = usinθ
Now recall from equations of motion that v = u + at
at maximum height, the final velocity v = 0
So 0 = u + at
but a = g(since its against gravity)
so gt = u
therefor
t = usinθ/g
t = usinθ/g
thats the time takes to reach maximum height
So the time to reach maximum height and then back to the gruond should be = 2t = T
so T= 2t = 2usinθ/g
Maximum Height (H_{max})
This is the highest attainable height of a projectaed body of initail velocity u. It can also be said to be the highest vertical distance reached by the projected body as measured from the horizontal projection plane.
We can derive this parameter by recalling that
v^{2} = u_{y}^{2} + 2as
at maximum height, the final velocity v = 0
so 0 = u_{y}^{2} + 2as
a = g, so that
2gs = u_{y}^{2}
by subject formular,
H_{max} = s = u_{y}^{2}/2g
recall that U_{y} = usinθ
so H_{max} = u^{2}sinθ/2g
Range (R)
This is defined as the horizontal distance from the point of projection to the point at which the projected body lands or hits the ground again.
We can derived this by considering the equations of motion that says
s = ut + (1/2)at^{2}
note that for range acceleration due to gravity is zero, i.e a = g = 0, and t = time of flight = T, and u = horizontal component of the initial velocity u
the horizontal component of the initial velocity u = u_{x} = ucosθ
so, s = u_{x} × T
s = (ucosθ × 2usinθ)/g
s = (u^{2}2sinθcosθ)/g
note from trig. that sin2θ = sin(θ + θ) = 2sinθcosθ
so Range R = s = (u^{2}sin2θ)/g
So from the above, one can obtain a maximum range when θ = 45_{o}
such that R = u^{2}/g.
(1)A shooter wants to fire a bullet in such a way that its horizontal range is equal to three times its maximum height. At what angle should he fire the bullet to achieve this?
A. 53^{o}
B. 68^{o}
C. 30^{o}
D. 45^{o}
From the question, the shooter wants
R = 3H_{max}
from projectile equations
R = ^{(u2sin2θ)}/_{(g)}
and H_{max} = ^{(u2sin2θ)}/_{(2g)}
hence ^{(3u2sin2θ)}/_{(2g)} = ^{(u2sin2θ)}/_{(g)}
cancelling out same parameters we have
that ^{(3sin2θ)}/_{2} = sin2θ
from trig, sin2θ = sin(θ + θ) = 2sinθcosθ
so that ^{(3sin2θ)}/_{2} = 2sinθcosθ
thats sinθ/cosθ = 4/3
recall that sinθ/cosθ = tanθ
then θ = tan^{1}(4/3) = 53.13^{o}
So Correct Option is A.(2) A hose of crosssectional area 0.5mms^{2} is used to dischrge water from a water tanker at a velocity of 60msms^{1} in 20s into a container. If the container is filled completely, the volume of the container is?
A. 600cm^{3}
B. 2400cm^{3}
C. 240cm^{3}
D. 600mm^{3}
Solution
recall that volume = cross sectional area × height
So our work here is to find height by using equation of motion
In here the volume of the water that flows out of the hose at time t is same volume that fill the tank. So the volume of water that flow from the hose at t= 20s
at velocity v = 60m/s and crosectional area = 0.5m is given thus
volume = area × height
but height = velocity v × time = 60 × 20 = 1200m, (i.e distance = velocity × time )
so that volume = 1200 ×0.5 = 600m^{3}
Correct Option is D
(3) A motorcyclist travelling at 30ms^{1} starts to apply his brakes when he is 50m from the traffic light that has just turned red. If he just reached the traffic light, his deceleration is?
A. 18ms^{2}
B. 10ms^{2}
C. 9ms^{2}
D. 5ms^{2} Solution The parameters given are
u = 30ms^{1}
s = 50m
50m is the distance between the traffic and were the driver saw it.
So from were it saw the traffic to were final velocity = 0, the deacceleration is given thus
since v = u + at; v = 0; a = u/t
but t = s/u = 50/30 = 5/3s(i.e time = distance/velocity)
then a = 30/(5/3) = 30 × 3/5 = 6 × 3 = 18m/s^{2}
So the correct Option is ACan you?
(1) An object is projected from a height 80m above the ground with a velocity of 40ms^{1} at an angle of 30^{o} to the horizontal. The time of flight is?
A. 16s
B. 10s
C. 8s
D. 4s
Updated Solution
recall that time of flight T = 2usinθ/g
u= 40ms^{1}, θ = 30^{o}, g = 10m/s^{2}
T = (2 * 40 *sin30)/10 = 4s
So answer is option D(2) What is the accceleration between two points on a velocitytime graph which has coordinate (10s, 15ms^{1}) and (20s, 35ms^{1})?
A. 1.75ms2
B. 3.50ms2
C. 1.00ms2
D. 2.00ms2
Updated Solution
acceleration is define as change in velocity over time
i.e a = (v^{2}  v^{1})/(t_{2}  t_{1})
a = (35  15)/(20  10) = 20/10 = 2.00ms^2
Answer is option D(3) A bullet fired vertically upward from a gun held 2.0m above the ground reaches its maximum height in 4.0s. Calculate its initial velocity.
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A. 10ms^{1}
B. 8ms^{1}
C. 40ms^{1}
D. 20ms^{1}
(g=10ms^{2}) Updated Solution
recall that the time to reach maximum height is t = (usinθ)/g
t = 4s, and θ = 90(since its vertically upward the angle it makes with the horizontal is 90)
so by subject formula, u = g * t = 10 * 4 = 40ms^{1}
So option C answer.
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