unit-1.4 uniformly accelerated motion in one dimension ... uniformly accelerated... · 1.4...

11th grade Fizx

Unit-1 Force & Motion

1.4 Uniformly Accelerated Motion in One Dimensionpage - 56

Solutions

Force & Motion Chapter-1

56

Problems

1. The velocity of a car is uniformly increased from 10 m/s to 20 m/s in 4 s. Calculate a) the acceleration of the car. b) the distance travelled by the car in 4 s. c) the distance travelled during the fourth second. 2. A car accelerates at 4 m/s2 from rest for 6 s. Then, the car moves with constant speed. Finally, the car decelerates at 8 m/s2 and comes to stop. If the total displacement of the car is 168 m, how many seconds does the car travel with constant speed?

3. Just a car accelerates at 2,5 m/s2 from rest; a bus moving at a speed of 20 m/s passes the car. How long does it take for the car to overtake the bus? 4. The driver of a car moving with 20 m/s saw the red light flashes on an intersection. The reaction time of the driver is 0,4 s and the car decelerates at a rate of 5 m/s2. Calculate the distance traveled by the car until it is stopped.

a) a=(v-v0)/t=(20-10)/4

a=2,5 m/s2

b) vaverage=(10+20)/2=15 m/s

x=15.4=60

c) The velocity of the car at 3 s v3=10+(3.2,5)=17,5 m/s

The velocity of the car at 4 s v3=10+(4.2,5)=20 m/s

The average velocity in 4th second vaverage=(17,5+20)/2=18,75 m/s

x=18,75.1=18,75 m

11th grade Fizx



Solutions


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Problems



Accelerating for 6s

vfinal=6.4=24 m/s

vaverage=(0+24)/2=12 m/s

x1=12.6=72 m

Moving at constant velocity

v=24 m/s

decelerating

t=(-24)/(-8)=3 s

x2 : displacement


x3=12.3=36 m

168=72+x2+36 x2=60 m

11th grade Fizx



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Problems



xcar=(1,25t).t=1,25.t2

for the car;

vfinal=2,5.t

for the bus;

xbus=20.t

xcar=xbus

1,25.t2=20t

t=16 s

vaverage=(2,5.t+0)/2=1,25.t

11th grade Fizx



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Problems



in reaction time;

x1=20.0,4=8 m

decelerating;

t=(0-20)/(-5)=4 s


x2=10.4=40 m

xtotal=8+40=48 m

11th grade Fizx



Solutions

Chapter-1 Force & Motion

57

5. Two trains move on the same rails in the same direction. The leading train travels at 10 m/s and the subsequent train travels at 40 m/s. When the distance between the trains is 90 m, the subsequent train begins to decelerate. What minimum deceleration is required to prevent a collision? 6. A motorcycle is 50 m behind a car. The two vehicles start from rest at the same moment. The car has a constant acceleration of 3 m/s2. The motorcycle passes the car after the car has travelled 150 m. a) How much time is needed for the

motorcycle to pass the car? b) What was the acceleration of the

motorcycle? c) What are the speeds of the vehicles when

the motorcycle passed the car?

7. In the 100 m race a sprinter accelerates uniformly for 8 s from rest to a maximum speed. After achieving the maximum speed he runs the remainder of the race without speeding up or slowing down. If the total race is run in 14 s, how far does he run during the accelerating stage? 8. A car accelerates uniformly from rest. Velocities of the car at points A, B and C are given in the figure.

If the distance between points A and B is 20 m, calculate the distance between points B and C.

vfinal=10 m/s for the subsequent train to prevent a collision.


for the subsequent train;

X1=25.t

for the leading train;

X2=10.t

X1=90+X2 25t=90+15t t=6 s

a=(10-40)/(6)=-5 m/s2

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b) for the car;

vfinal=3t

vaverage=(0+3t)/2 vaverage=3t/2

vaverage=(0+3t)/2 vaverage=3t/2

Xcar=(3t/2).t

150=3t2/2

t=10 s

a) for the motorcycle;

Xmotor=(vmotor/2).t

vmotor=40 m/s

amotor=40/10=4 m/s2

c) vmotor=40 m/s

vcar=30 m/s

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accelerating;

vfinal=8a

vaverage=(8a+0)/2=4a

x1=4a.8=32a

x2=8a.6=48a

Moving with constant velocity;

xtotal=32a+48a=100 a=5/4 m/s2

x1=32a=32.(5/4)=40 m

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Solutions


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The car moves with a constant acceleration

tAB=t then tBC=t

vaverage-AB=(v+3v)/2=2v

20=2V.t V.t=10

vaverage-BC=(3v+5v)/2=4v

xBC=4V.t=4vt=40 m

xAB=vaverage-AB.t

The changes in velocities of the object from A to B and from B to C are the same. So,

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9. Acceleration-time graph of a car is given. The initial position of the car (x0) is -20 m and the initial velocity of the car (v0) is +4 m/s.

Plot velocity-time and position-time graphs of the car.

10. Position-time graph of a car accelerating uniformly is given. The velocity of the car is zero at t=2 s.

Plot velocity-time and acceleration-time graphs of the car.

Δx1=((4+10)/2).2=14 m

Δx2=((2+10)/2).4=24 m

Δx3=((2+12)/2).2=14 m ΔV1=3.2=6 m/s

ΔV2=-2.4=-8 m/s

ΔV3=5.2=10 m/s

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9. Acceleration-time graph of a car is given. The initial position of the car (x0) is -20 m and the initial velocity of the car (v0) is +4 m/s.

Plot velocity-time and position-time graphs of the car.

10. Position-time graph of a car accelerating uniformly is given. The velocity of the car is zero at t=2 s.

Plot velocity-time and acceleration-time graphs of the car.

displacement is 4 m and it is decelerating

displacement is -4 m and it is accelerating

4=(v1.2)/2 then v1=4 m/s

-4=(v2.4)/2 then v2=-2 m/s

a1=((0-4)/2)=-2 m/s2 a2=((-2-0)/4)=-0,5 m/s2

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Solutions


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11. Velocity-time graphs of two cars K and L are given. They are at the same position at t=0. The distance between the cars is 20 m at time “t”.

What is the distance between the cars at time “3t”? 12. A car starts from rest and its acceleration-time graph is given. The displacement of the car in t-2t interval is “x”.

What is the displacement of the car in 2t-3t time interval? (The initial position of the car is zero.)

13. The velocity-time graphs of cars K and L are given. Car K is 25 m in front of car L at t=15 s.

What is the position of car K with respect to car L at t=0? 14. An object decelerates uniformly and stops at 2t seconds.

If the average velocity of the car is -20 m/s, calculate “t” given in the graph.

For 0-t time interval; xK=X and xL=-X The distance between them at "t" is 2x=20 then x=10 m.

For 0-3t time interval; xK=4X and xL=3X The distance between them at "3t" is 4x-3x=x=10 m.

Let’s say this area is equal to “x”.

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Solutions


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ΔV1=2at ΔV2=at ΔV3=0

X=5at2/2 then at2=2x/5

x/=3at2 then x/=6x/5

The distance between them is 25 m at t=0.

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The area of v-t graph gives the displacement.

The displacement of car K in 0-15 s interval;

xK=(10.5/2)+((10+20)/2).5+(20.5)

xK=25+75+100=200 m

The displacement of car L in 0-15 s interval;

xL=((10+20)/2).5+(10.5)++((10.5)/2)

xL=75+50+25= 150 m

Ki Li Kf Lf

150 m200 m

25 m25 m

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X=vaverage.t

(-v0.2t)/2=20.2t

v0=-40 m/s

-160=(-40.2t)/2

t=4 s

0-t interval; x1=-120 m

t-2t interval; x2=-40 m

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15. A freely falling stone covers 80 m in the last 2 s of its motion. Calculate a) the time of flight of the stone. b) the height that the stone was released with

respect to the ground. c) the speed of the stone just it hits the

ground. (Ignore friction and g=10 m/s2.) 16. A ball is released from point A. The time needed to cover the distance between points A and B is “2t” and the time needed to cover the distance between B and C is “3t”.

What is the ratio of h1 to h2? (Ignore friction.)

17.

An object is thrown vertically from a height of 80 m with a speed of 30 m/s. What is the height of the object from the ground after 1 second? (Ignore friction and g=10 m/s2.)

18. Three positions and the speeds at those positions of a ball falling vertically are given in the figure. The time needed to cover the distance between points A and B is “t1” and the time needed to cover the distance between B and C is “t2”.

a) What is the ratio of t1 to t2? b) What is the ratio of h1 to h2? (Ignore friction.)

The speed of the stone just it hits the ground is “v=10t” where “t” is the time of flight of the

object.

The speed of the stone before 2 s it hits the ground was “10t-20”

h=vave.t 80=((1ot+10t-20)/2)*2

t=5 s

height that the stone was released (h) h=(10.52)/2=125 m

V=10t=10.5 V=50 m/s

a)

b)

c)

11th grade Fizx



Solutions


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17.





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15. A freely falling stone covers 80 m in the last 2 s of its motion. (Ignore friction and g=10 m/s2.) Calculate a) the time of flight of the stone. b) the height that the stone was released with


ground. 16. A ball is released from point A. The time needed to cover the distance between points A and B is “2t” and the time needed to cover the distance between B and C is “3t”. (Ignore friction.)

What is the ratio of h1 to h2?

17. An object is thrown vertically from a height of 80 m with a speed of 30 m/s. (Ignore friction and g=10 m/s2.)

What is the height of the object from the ground after 1 second? 18. Three positions and the speeds at those positions of a ball falling vertically are given in the figure. The time needed to cover the distance between points A and B is “t1” and the time needed to cover the distance between B and C is “t2”. (Ignore friction.)

a) What is the ratio of t1 to t2? b) What is the ratio of h1 to h2?

ADVANCED PHYSICS

2t

3t2gt

5gt

h1 =0+2gt

2.2t=2gt2

h2 =2gt+5gt

2.3t= 21gt2

2

⎫

⎬

⎪⎪⎪

⎭

⎪⎪⎪

⇒

h1

h2= 2gt2

21gt2

2

= 421

g

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ADVANCED PHYSICS

40 m/s

1 s h

h= 30+402

⎛⎝⎜

⎞⎠⎟ .1=35 m

from the ground;

80 - 35 = 45 m

g

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Solutions


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17.





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ADVANCED PHYSICS

g

t1

t2

a)2v - v=g.t1

3v - 2v=g.t2

⎫

⎬⎪⎪

⎭⎪⎪

⇒ t1 = t2

b)

h1 =v+2v

2.t1

h2 =3v+2v

2.t2

⎫

⎬

⎪⎪⎪

⎭

⎪⎪⎪

⇒ h1

h2= 3

5

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Solutions


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19. A ball is thrown vertically upward from the ground with an initial speed of 50 m/s. Calculate a) the time of flight of the ball. b) the maximum height reached by the ball

from the ground. c) the velocity of the ball from the ground

after 7 s. (Ignore friction and g=10 m/s2.) 20. A stone is thrown vertically upward from a height of 40 m with an initial speed of 10 m/s. Calculate a) the time of flight of the stone. b) the velocity of the stone after 4 s. c) the height of the stone from the ground

after 3 s. (Ignore friction and g=10 m/s2.)

21. Two objects X and Y are thrown vertically as shown in the figure. The time of flight of object X is “2t” and the time of flight of the object Y is “t”.

What is the ratio of h1 to h2? (Ignore friction.) 22. A balloon rises up at a constant speed of 8 m/s. The passenger throws a ball vertically upward at 4 m/s with respect to balloon. The ball hits the ground after 5 s. What was the height of the balloon from the ground at the instant the passenger had thrown the ball? (Ignore friction and g=10 m/s2.)

a)

tup = 5010

= 5 s ⇒ t flight = 10 s

b)

hmax = 50+02

⎛⎝⎜

⎞⎠⎟ .5 = 125 m

c)v = 50 - 10.7( )=-20 m / s

11th grade Fizx



Solutions


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a)

h= vi.Δt+ 12

.g.Δt2

-40= 10.t - 12.10.t2 ⇒ t2 - 2t+8=0

t = 4 s

b)v f = vi +gΔt

v = 10+(-10).4 ⇒ v =-30 m / s

c)

h= vi.Δt+ 12

.g.Δt2

h= 10.3 - 12.10.32 = −15 m

from the ground;40 - 15 = 25 m

11th grade Fizx



Solutions


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61


19. A ball is thrown vertically upward from the ground with an initial speed of 50 m/s. (Ignore friction and g=10 m/s2.) Calculate a) the time of flight of the ball. b) the maximum height reached by the ball


after 7 s. 20. A stone is thrown vertically upward from a height of 40 m with an initial speed of 10 m/s. (Ignore friction and g=10 m/s2.) Calculate a) the time of flight of the stone. b) the velocity of the stone after 4 s. c) the height of the stone from the ground

after 3 s.

21. Two objects X and Y are thrown vertically as shown in the figure. The time of flight of object X is “2t” and the time of flight of the object Y is “t”. (Ignore friction.)

What is the ratio of h1 to h2? 22. A balloon rises up at a constant speed of 8 m/s. The passenger throws a ball vertically upward at 4 m/s with respect to balloon. The ball hits the ground after 5 s. (Ignore friction and g=10 m/s2.) What was the height of the balloon from the ground at the instant the passenger had thrown the ball?

ADVANCED PHYSICS

g

t t

t

v=gt

vf-2v=gt

vf=3v

h1 =0+v2

.t= v.t2

h2 =2v+3v

2.t= 5v.t

2

⎫

⎬

⎪⎪

⎭

⎪⎪

⇒ h1

h2= 1

5

11th grade Fizx



Solutions


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8 m/s

4 m/s

hg

The velocity of ball wrt ground is 12 m / s.

h= vi.Δt+ 12

.g.Δt2

-h=(12.5)- 12.10.5 2 ⇒

h=65 m

ground

(wrt balloon)

unit-1.4 uniformly accelerated motion in one dimension ... uniformly accelerated... · 1.4...

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