## Problems Linear motion Content

## Problems Linear motion

__Problems Linear motion__

In this post and in few of my posts to come, I would like to solve problems on linear motion,freely falling bodies,vertically projected up bodies and projectiles .

1.An object accelerates from rest to a velocity 20m/sec in 4seconds.If the object has uniform acceleration, find its acceleration and displacement in this time.

Soln: From the data given in the problem we have,

Initial velocity = u =0,

final velocity v=20 m/sec,

Time of journey t=4sec,

Acceleration a = = = 5

Displacement S = ut + = 0 (4)+

S = 40m.

2.An object starting from rest moves with uniform acceleration of 3 for 6sec.Find its velocity and displacement after 6seconds.

Soln: From the data given in the problem we have,

Initial velocity of the object u = 0,

Acceleration of the object a= 3,

Time of journey t =6sec.

Final velocity of the object v=u+at =0 + 3(6) = 18 m/sec.

Displacement S= ut + = 0 (6)+

S= 54m.

3.An object starting from rest moves with uniform acceleration of 4.Find its displacement i) 5seconds ii) in 5th second iii) 8th second.

Soln : From data given in the problem

Initial velocity of the object u=0,

Acceleration of the object a=4,

i) time t=5 seconds,

Displacement of the object S=ut + = 0 (5)+

Displacement of the object in 5seconds S= 50m.

ii) Displacement of the body in 5th second =?

Let us substitute n=5 in the formula =u+a(n-1/2)

=0+4(5-1/2) = 4(4.5) =18m.

iii)Displacement of the body in 8th second =?

Let us substitute n=8 in the formula =u+a(n-1/2)

=0+4(8-1/2) = 4(7.5) =30m.

4.An object started moving with an initial velocity of 10m/sec, after traveling a distance of 5m gets a velocity 20m/sec.Find its i) acceleration ii) time taken for 5m displacement.

soln: From the data given in the problem,

Initial velocity of the object u=10m/sec,

Final velocity of the object v= 20m/sec,

Displacement S=5m,

i) acceleration a=?

Substitute the values of u,v and S in the equation =2as,

we get =2a(5)

300 = 10a or a = 300/10=30.

ii)Time t=?

Substitute the values of u,v and a in the equation v=u+at,

we get 20=10+(30)t ; 10=30t

t =10/30 = 1/3 = 0.333 sec.

5.When an observer started observing a car it’s velocity was x m/sec , if it travels for 10sec with uniform acceleration 2.5and its velocity increases to 75m/sec.Find i) Initial velocity of the car ii) Displacement of the car in 10sec iii) Displacement of the car in first 5sec and last 5sec, what is your inference.

Soln: From the data given in the problem,

Initial velocity of the car u = x (say),

Final velocity = 75m/sec,

Acceleration a = 2.5,

Time of journey t=10sec.

i) Substitute the values of u,v,a and t in the equation v=u+at

we get 75 = x+2.5(10) ; x=50 m/sec.

ii) Let the displacement of the car in 10sec be S

Substitute the values of u,a and t in the equation S=ut +

We get S = 50(10)+ ;

S = 500+125 ; s=675 m

ii) Let the displacement in first 5sec be

Substitute t=5 sec and thve values of u and a in the equation =ut +

we get =50(5) +

= 250 + 31.25 = 281.25 m.

Let the displacement in next 5sec be

= S –

= 675 – 281.25 = 393.75 m

We can observe that, even though the time of journey is same , >

Displacement of the body in second half is greater than in the first half time of its journey.

6 .A cheetah can accelerate from rest to 24.0 m/s in 6.70 s.

Assuming constant acceleration, how far has the cheetah run in this time?(Question by suzi in yahoo answers).

Soln: From the data given in the problem,

Initial velocity of the cheetah u=0,

Final velocity of cheetah v=24 m/sec,

Time t = 6.7sec

Distance traveled by cheetah be S=?

The equations of motion are = 2aS- – – – – (1)

and a = – – – – – (2)

From equations (1) and (2) we get =2S

simplifying it we get S = =

Substitute the values of u,v and t we get S = =12(6.7)=80.4 m

7) A car covers first halt of the distance between two places at a speed of 30 km/hr and the second half at a 90km/hr.What will be the average speed of the car?

Soln:

Method I: Let the total distance between the places be S.

Time taken to cover First half = = hours.

Time taken to cover Second half = = hours.

Total time t = + =+= = .

Average speed = Total distance /total time.

= S/t =

= 45km/hr.

Method II (Short cut): If the body covers 1st half of distance with a speed x and the second half with a speed y,then the average speed = .

Average speed = latex frac{2700}{60}$ = 45km/hr.

8) A body starts from rest and acqires a velocity of 400m/sec in 10seconds.Calculate the acceleration and distance traveled.

Soln: From the data given in the problem

Initial velocity of the body u=0,

Time of journey t=10sec,

Final velocity v=400m/sec,

Acceleration a=? and distance traveled in 10sec s=?

Substitute the values of u,v and t in the equation v=u+at,

we get 400= (0)(10) + a (10) ; 10a=400

a= 40

Substitute the values of u,a and t in the equation S= ut +

S=(0)(10)+ = 0+2000 = 2000m.

9) A body moving with uniform acceleration covers 6m in second and 16m in second.Calculate the initial velocity,acceleration and distance moved in second .

Soln: From the data given in the problem

Distance moved in second = 6m,

Distance moved in second = 16m,

– = a(4-2) =2a

therefore 2a = – =16-6 =10

a = 5 m/

Substitute the values of ,a and n=2 in the equation =u+a(n -1/2)

we get 6=u+5(2-1/2) ; 6=u+7.5

u=-7.5 +6 = -1.5 m/sec.

**Distance moved in the second = u+a(6-1/2),**

Substitute the values of u,a in the above equation we get second=-1.5+5(5.5)

second =-1.5+27.5 =26m.

10) An object started traveling with a velocity 2m/sec moves with an uniform acceleration of 3 m/.

i) Find the ratio of displacements in a) ,, seconds b) ,, and seconds .

ii)Find the ratio of velocities a) ,, seconds b) ,, seconds .

Soln: From the data given in the problem

Initial velocity u=2 m/sec,

Acceleration a =3 m/,

i- a) The ratio of displacements in ,, seconds

From the formula :: = (2u+a) : (2u+5a) : (2u+9a)

:: = (4+3) : (4+15) : (4+27)

:: = 7:19:31 .

i-b) The ratio of displacements in ,, seconds

From the formula :: = (2u+3a) : (2u+7a) : (2u+11a)

:: = (4+9) : (4+21) : (4+33)

:: = 13 : 25 : 37 .

ii -a) From the formula ::: . . . . . . . . := (u+a) : (u+2a) : (u+3a) : . . . . . . . . . : (u+na).

The ratio of velocities a) ,, seconds

:: = (u+a) : (u+3a) : (u+5a) = 5 :11 : 17 .

ii- b) The ratio of velocities a) ,, seconds

:: = (u+2a) : (u+4a) : (u+6a) =8 :1 4 :20 .

11) A body travels 200cm in the first two seconds and 220cm in the next four seconds.What will be the velocity at the end of the seventh second from the start?

Soln: The displacement of the body in first 2 sec =200cm,

Let the initial velocity = u(say) ,

Acceleration = a(say), time =2sec

= +,

Substitute the value of in above equation, we get

= 2u+,

= 2u +2a ; 2(u+a) =200

Therefore u+a = 100 – – – – – – – – – – – – – – – – – – – (1)

Given that the body travels 220cm in next 4sec.That is from the start it displaces 200+220 = 420 cm in 6sec.

Displacement =420 cm,

time =6 sec,

substitute theses values

in the equation = u+,

= 6u+=6u+18a,

6(u+3a) = 420 ; u+3a = 70 – – – – – – – – – – – – – – – – – – – (2)

Solving equations (1) and (2) or Eq (2) – Eq(1)

we get 2a= -30 ; a=-15 cm/,

Substitute value of a in Eq(1) we get u-15 = 100,

u = 115 cm/sec.

The velocity at the end of second v=u+a

we get v= 115+(-15)(6) =115-90 =25cm/sec.

Therefore final velocity v=25cm/sec.

12.A subway train starts from rest at a station and accelerates at a rate of 16.5 for 13.1 sec.It runs at constant speed for 69.7s and slows down at a rate of 3.45 until it stops at the next station.What is the total distance covered?

**Soln:** From the data given in the problem,

Initial velocity of the train u = 0 m/sec,

Acceleration a = 16.5 ,

time t = 13.1 sec,

Velocity after 13.1 sec v=?

and the distance traveled =?

Substitute the values in the equation v =u+at

we get v= 0+(16.5)(13.1) = 216.15 m/sec.

Substitute in equation = ut +

we get = (0)(13.1)+

= 0+1415.78 =1415.78 m

After that the train travels with velocity v=216.15 m/sec for 69.7sec. calculate distance traveled during this time.

v=216.15 m/sec, t = 69.7 sec =?

substitute the values in equation = vt = (216.15)(69.7)=15065.66 m

Finally the trains decelerates at the rate of 3.45 and comes to rest.find distance traveled before coming to rest.

Acceleration a = -3.45 ,

Initial velocity u = 216.15m/sec,

Final velocity v=0,

distance traveled =?

substitute the values in = 2as,

we get = 2(-3.45)

-6.90 = -46720.82

= 6771.13.

Total distance traveled by train S = ++ = 1415.78+15065.66+6771.13=23252.57 m or 23.252 km.

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