# 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 $6^{th}$ 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.