Linear motion, velocity time graph, motion under gravity
SPEED, DISPLACEMENT, VELOCITY, ACCELERATION
Velocity time graph
Motion under gravity
In this blog I will be solving multiple questions with answers on speed , displacement, velocity and acceleration
Speed
SPEED is defined as the rate of change of distance with respect to time. It is also know as average speed.
Speed is a scalar quantity. It has a magnitude but no direction. The unit of speed is meter per second (m/s)
Average speed = ∆distance/∆time =
Final distance-initial distance/final time-initial time.
Average speed = distance/time =s/t
QUESTIONS AND ANSWERS ON SPEED
1.1A man walks a distance of 58km in 2hours calculates his average speed.
Average speed =distance/time
=58km/2hrs X 1000m/3600s X hrs/km
=8.06m/s
Hint: 1km= 1000meters
1hours =3600seconds
1hours=60minutes
1minutes=60seconds
1.2 A taxi driver with a speed of 20km/hrs saw a goat on a highway and applied a break. It came to rest after 10seconds. calculate the distance covered before coming to rest.
Solution
Speed, S =20km/hrs
Time,t=10seconds.
First step
Convert 20km/hrs to m/s
Speed =20x1000/3600
Speed=5.56m/s
Speed=distance/time
Make distance subject of the formula
Distance= speed x time
= 5.56x10 = 55.6m
1.3 A heavy-duty truck driver traveling at a speed of 182km/hrs received a phone call from his wife on his cell phone. How far is he, in kilometers, in 10seconds later, when he received the call from his wife.
Solution
Speed, S = 182km/hrs
Time,t=10seconds
Distance =speed x time
= 182km/hrs x 10hrs/3600 =0.51km
1.4 the time rate of change of distance is known as ------
(A) speed
(B) velocity
(C) acceleration
(D) Displacement
Answer A
1.5 which of these is a scalar quantity?
(A) momentum
(B) Force
(C) speed
(D) acceleration
Answer C
1.6 Juliet run to the kitchen when she perceived a small of burning food with a speed of 18m/s. If the distance between where she was sitting down to kitchen is 25m. How long does it take her to reach the kitchen?
Solution
Speed, S=18m/s
Distance, D = 25m
Time,t =
Speed = Distance/time
Time =25M/ 18MS⁻¹
T=1.39seconds
1.7 A tricycle is at a position X = 20m at a time t = 10s and later change position X = 50m at a time t = 25s.what is the average speed?
Solution.
X₁=20m
X₂=50m
t₁=10s
t₂=25s
Sₐᵥ=
Sₐᵥ=∆x/∆t
=50-20/25-10
=30/15 = 2ms⁻¹
DISPLACEMENT
This is defined as the change in the position of an object, particle, or body in a specific direction. Displacement has a magnitude and direction thus it is a vector quantity.
QUESTION ON DISPLACEMENT
2.8 A man entered a retail shop to buy some snacks, after he bought the snacks he moved 5km North to his girlfriend's house after that he also move 20km towards the South. what is his displacement?
Solution
Displacement =change in distance
Displacement=∆X =X₂ – X ₁
Displacement =20km-5km =15km
2.9 A car is at X = 10m and is later seen at X =34m .find the displacement.
Solution
X₁= 10
X₂=34
Displacement=∆x = 34-10 = 24km
VELOCITY
Velocity is the rate of change of displacement with time. It has a magnitude and direction so it is a vector quantity.
Uniform velocity is defined as the velocity of a moving particle in which equal distances are covered in equal time intervals, regardless of how small the time intervals may be. that’s an increase in velocity concerning an increase in time.
VELOCITY = displacement/time
=Change in displacement/change in time
=Final displacement-initial displacement/Final time -initial time
Hint: Speed is used in place of velocity and vice versa. velocity and speed have the same S.I unit (m/s). The difference between velocity and Speed is that:
Velocity has a magnitude and direction (vector quantity)
While speed has only magnitude no direction (a scalar quantity)
RECTILINEAR ACCELERATION
This is a term used to describe the acceleration of a body on a straight line.
Rectilinear acceleration is defined as the rate of increase or change of velocity along a straight-line path in a unit of time. When the velocity of a particle changes, it could be either it is undergoing acceleration or deceleration.
DECELERATION
It is defined as the decreasing rate of change of velocity with time.
Acceleration =change in velocity/Time taken for the change
EQUATION OF UNIFORMLY ACCELERATED MOTION.
FIRST QUESTION OF MOST
a=change in velocity/Time interval
a=v-u/t
The second equation of motion
S=ut +1/2at²
Third equation of motion
V²=u²+2as
Hint;
v—--final velocity
u----initial velocity
a-----acceleration/deceleration
S-----distance
t----time
QUESTIONS ON VELOCITY AND ACCELERATION
3.10 A Taxi driver traveling along a road that passes through Forest at 180km/hrs sees an elephant crossing the road and immediately applies his brakes and comes to a stop with uniform deceleration in 35seconds .what distance does the car travel after the brakes were applied.
Solution
First step
convert 180km/hrs to m/s
180km/hrs = 180X1000m/60X60s
= 50m/s
Initial velocity,u = 50m/s
Second Step
Find the deceleration/retardation
Using this formula
Retardation a= v-u/t
Aa= 0- 50/35 = -1.43m/s²
Hint the negative sign (-1.43m/s) shows a retardation/deceleration.
Third step
Using any of these equations of linear motion to find the distance
i.S=ut+1/2at²
Or
ii. V²=u²+2as
I will use the two-equation to solve the problem and still get the same answer
i.S=ut+1/2 at²
u= 50m/s
a=1.43m/s²
t=35seconds
S=ut+1/2at²
S= 50X35 + ½ x(- 1.43) x 35²
S=1750-875.875
S=874.125m
Or
ii. V²=u²+2as
V=0
Making S the subject of a formula
S= u²/2a
S= 50²/2x1.43
S= 2500/2.86
S=874.125m
3.11 A body starts from rest and accelerates uniformly at 50ms⁻².calculate the time taken by the body to cover a distance of 1km.
Solution
a=5ms⁻²
s=1km=1000m
v ²=u²+2as
U=0
Hint: Reason why u =0
The body accelerates from rest.
V²=2x5x1000
V²=10000
V=√10000
V=100
a=v-u/t
make t subject of a formula
t=v-u/a
t=100-0/5
t=20m/s
3.12 A particle starts from rest and acceleration uniformly 25m/s².The distance covered in 15seconds is -------
Solution
Hint the body start from rest and accelerate uniformly that mean
Initial velocity u=0
Step one
Find the final velocity
a=25m/s²
t=15seconsa
U=0
V=?
a= V-U/t
V=at+u
V=50 x15 + 0
V=750
Step two
Find the distance using the third equation of linear motion
V²=u²+2as
750²=2x50xS
562500/100 =S
S= 5625m
Converting it to kilometer
S=5.6 x 10³km
3.13 How long will it take a particle to cover a distance of 40m if uniformly accelerated from rest at the rate of 2m/s².
Solution
T=?
a=2m/s²
S=40m
V=at+u
V=2t+ 0
V=2t
V²=u²+2aS
(2t)²= 0+ 2x5x40
4t²/4 = 400/4
t²= 100
t = √100
t=10s
3.14 A car starts from rest and moves with a uniform acceleration of 6m/s² until it attains a velocity of 25m/s. calculate the time taken to attain this velocity
Solution
V=at+u
25=6t+0
26/6 = 6t/6
T=4.17s
3.15 A train starts from rest and moves with a uniform acceleration of 14m/s² .what distance does it cover in the last dirty second?
Solution
The time taken to cover the distance S IS a time Interval between 5th and 4th
S=S₅-S₄
U=0
S =1/2a(t²₅ – t²₄)
S=1/2 x 14(5²-4²)
S = 7x(25-16)= 7x9
S=63m
3.16 A particle accelerates uniformly from rest at 5m/s².calculate its velocity after traveling 5m
Solution
a= 5m/s²
Distance s=5m
U=0
V=?
V²=u²+2as
V²=0²+2x5x5
V²=100
V=√100
V=10m/s
3.17 An airplane lands on a runway at a speed of 69.44m/s and covered a distance of 347m before it was brought to rest. How long does it take the airplane to come to rest?
Solution
Initial velocity u= 250km/s = 250000/3600 = 69.44m/s
Final velocity V=0
Retardation, V²= u²+2aS
0²=u²+2aS
-u²=2aS
a = -u²/2S
a=- (69.44)²/2x347 =-6.948m/s²
Time t v= at+u
V=0
0=-6.948 x t + 69.44
-69.44/-6.948=-6.948t/-6.948
9.99=t
t=10s
3.18 After the application of the break, a truck with a speed of 180km/hrs was brought to rest in 20seconds. Calculate the distance covered by the truck after the brakes were applied.
Solution
Initial velocity, u= 180km/hrs
Convert to meter per second
U= 180x 1000/3600 = 50m/s
Time,t = the 20s
a= v-u/t = 0-50/20 =-2.5m/s²
S= ut+1/2at²
S=50x20+1/2(-2.5)x20² = 500m
S=500m
3.19 A particle is brought to rest from a velocity of 30m/s in 45seconds. calculate the retardation.
Solution
U= 30m/s
t=45seconds
V=0
A=?
a=v-u/t
a=0-45/30
a=1.5m/s²
3.20 A body accelerates uniformly from rest at 4m/s². Calculate its velocity after traveling a distance of 15m
Hint whenever a body accelerates from rest initial velocity u=0
a=4m/s²
Distance, s=15m
Initial velocity u=0
Final velocity V=?
Using the third equation of motion
V²=u²+2aS
V²=0²+2x4x15
V²=120
V=√120
V=10.95m/s
3.21 The Coordinates (15s,30m/s)
And (the 60s,70m/s) are two points on a velocity-time graph. Calculate the mean acceleration between the two points
Solution
t1=15seonds
t2=60seconds
V1=30m/s
V2=70m/s
aₘₑₐₙ=v₂₋ v₁/t₂-t₁
aₘₑₐₙ =70-30/60-15 =40/45
aₘₑₐₙ=0.88ms⁻²
VELOCITY TIME GRAPH
The graphical solution to problems in rectilinear acceleration.
VELOCITY time graphs are mostly used to solve problems in rectilinear acceleration and could be in different shapes depending on whether the velocity is increasing, decreasing, or constant.
Questions and answers
4.22 A Bugatti Veyron runs at a constant speed of 600m/s for 400seconds and then accelerates uniformly to a speed of 950m/s throughout 50seconds. This speed is maintained for 200 seconds before the Bugatti Veyron was brought to rest with uniform deceleration in 120seconds. Draw a velocity-time graph to represent the journey described above. From the graph:
i. Find acceleration while the velocity changes from 30m/s to 75m/s
ii. Find the total distance traveled in the time described
III. Find the average speed over the time described.
Solution
Acceleration
a=slope BC= Y₂ - Y₁/X ₂– X₁
a= 950-600/450-400
a=350/50 =7ms⁻²
Ii. In a velocity-time graph, the total distance Is equal to the area under the velocity-time graph.
AREA OABCDE = area of rectangle ABHO (A1) + area of trapezium BCGH (A2) + area of rectangle CDGF (A3) + area of the triangle DFE (A4)
Area ABHO = Distance covered during the constant speed of 600m/s for 400 seconds.
Area BCDH= distance covered during velocity change from 600m/s to 950m/s
Area CDFG = distance covered by the Bugatti Veyron during its constant velocity.
Area DEF = distance covered during retardation.
A1= OA X OH = 600x400=240km
A2= ½(BH +CG)HG
= ½(600+950) x 50 = 38.75km
A3= CD X DF = 950x200 = 190km
A4=1/2(FE x DF)
= ½(120 x 950) = 57km
Total distance = A1 + A2 + A3 + A4 = 240km + 38.75km +190km + 57km
Total distance = 525.75km
Note: the answers were converted to kilometers by dividing them by 1000.
iii. Average speed
= total distance traveled/total time taken
Convert 770second to an hour so that answer will be in kilometer per hour
770/60x60 = 0.2138hr
Average speed = distance /time
= 525.75/0.2138 = 2459.07km/hrs
4.23 A particle started at rest and move with an initial acceleration of 15m/s² for 45second. The acceleration was reduced to 25m/s² for the next 15second. The particle maintained the speed attained for 80second after which it is brought to rest in 10second. Illustrate the journey on a velocity-time graph.
(bi) calculate speed attained during the motion
ii. Calculate total distance traveled during the time of 45 second
iii. Average retardation as the body is being brought to rest
Iv. Average speed during the time 60second
Solution
Time t= 45s
Acceleration a=15m/s
Initial velocity u= 0
Final velocity,v = u+at
V= 0+15x45 =675m/s
The retardation started from a speed of 675m/s therefore initial velocity u=675m/s
Retardation, a =-25m/s²
The negative sign shows there was a reduction in acceleration (retardation).
V=675+(-25x15)
V= 675-375 = 300m/s
(I)The maximum speed attained is 675m/s
ii. Total distance covered during the first 45 seconds = Area of BOH = area of triangle = ½(BH x HO)
= ½ x 675x45 = 15187.5 m = 15.19km
iii. Average retardation = slope of DE = Y2 - Y1/X2-X1
= 300-0/140-150
Average retardation = -300/10
= -30m/s²
Note: the negative sign indicates retardation.
iv. Average speed during the first 60s = total distance in 60s/total time taken
Total distance = area of triangle BOH (A1) + Area if trapezium BCGH (A2)
A1=15.19km from ii solution
A2 = ½( CG + BH )HG
A2=1/2(300+675)x15 = 5212.5m
=5.21km
Average speed = 5212.5+15187.5/60 = 340m/s
MOTION UNDER GRAVITY: GRAVITATIONAL ACCELERATION.
When an object is thrown upwards or release from a height of h, its motion is under influence of Earth's gravity. An object is thrown upwards experiences a negative (-g) because its motion is in opposite direction to the gravitational pull of the Earth. On the other hand, when an object falls downwards it release from a height of h, it experiences a positive acceleration (+g) because its motion is in the same direction as that of the Earth's gravitational pull.
The equation of gravitation is derived by modification of rectilinear motion equation
V= u±at
S=ut±1/2at²
V²=u²±2aS
FOR BODY THROWN UPWARDS, THE FOLLOWING EQUATION S HOLD.
V= U-gt
H= ut -1/2gt²
V²=u²-2aS
FOR A BODY FALLING DOWNWARDS OR RELEASED FROM A HEIGHT, THE FOLLOWING EQUATIONS APPLY.
V=Ut+gt
H=ut + 1/2gt²
V²=u²+2gh
QUESTIONS AND ANSWERS
4.24 During house construction, a hammer fell from the top of the story building with an initial velocity of 45m/s If the hammer took a total time of 4second to reach the ground level, calculate the height of the story building. (g=10m/s²)
Solution
Velocity u= 0
Hint the body is falling a height of h
H= of ut + 1/2gt²
H=0x4 + 1/2x10 x4²
H= 80m
5.25 A body is projected vertically upwards fratoatopof a mountain with a speed of 85m/s². Determine the speed at its maximum point reached.
Solution
Answer is 0.0m/s
At a maximum height, the velocity of a particle projected upward is zero. The body is about to return to the ground.
5.26. A piece of metal drops from the top of a tower of a height of 50m .find the time it will take to reach the ground.
Solution
H=50
VELOCITY v²= u² +2gh
V²=2x10x50
V=√1000
V=31.62m/s
V=u+at
31.62=10t
T=3.16s
5.27 A pawpaw dropped to the ground from top of a pawpaw tree 75m tall. Calculate the time it will take to reach the group.
(g=10m/s²)
Height h=75m
Acceleration due to gravity g=10m/s²
T=?
Initial velocity u =0 (the body is falling)
Applying the second equation of motion
S= ut + 1/2gt²
75= 0xt + ½ x 10 xt²
75x2/10 = t²
150/10 = t²
t=√ 15
t= 3.87s
5.28 A mango drops to the ground from the top of a tree h tall. If the body land on the ground after 3seconds.I find the height h.
(ii) what speed did the mango use to fall from the height to the ground.
Solution
Time t= 3s
U= 0
g=10m/s²
H=?
V=?
h= ut + 1/2gt²
h= 0xt + ½ 10 x 3²
h= 45m
V= u + at
V= 0 + 10 x 3 = 30m/s
Or
V²= u² +2gh
V²= 0²+2x10x45
V²= 900
V=√900 =30m/s
5.29 A particle moving with an initial velocity u and accelerates until it attains a velocity of V within time t.The distance, s covered by the particle is given by the expression.
A. S=(v-u/2) t
B. S=(v+2u/2)t
C.S= (v+u/2)t
D.S=(2v-u/2)t
Answer C
5.30 which of these is not a vector quantity
A.velocity
B.Acceleration
C. Displacement
D.speed
Answer D
Speed has magnitude without direction (a scalar quantity)while other quantities have magnitude and direction (vector quantity )
EXERCISES
A particle at rest is given an initial acceleration of 10m/s² for 50seconds after which the acceleration is reduced to 6.0m/s² for the next 16seconds. The particle maintained the speed attained for 35 seconds then it come to rest after 20 second
Draw the velocity-time graph of the motion from the velocity-time graph calculate.
The maximum speed attained and reduction in velocity during the motion
Total distance traveled during the first 66second.
Average speed during the same time interval as in (ii) above
A bullet fired from a machine gun vertically upward from 145m above the ground, reaches its maximum height in 10second. Calculate (i) its initial velocity (ii) The maximum height attain as measure from the ground g=10m/a)
A carpenter mistakenly drops his hammer from a height of 34m. How long does the hammer take to hit the ground? (g=10m/s²)
Prove that F= m(V-u/t)
Differentiate between velocity and speed and similarities to them.
Do the exercises and post your work with your answer so that I can see it. If you are confused with any of the solution or exercise please don’t hesitate to contact through the comment box.
Comments
Post a Comment