Linear Motion Problems

Liner motion problems

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.

Solution: 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 = frac{(v-u)}{t} = frac{(20-0)}{4}= 5 msec^{-2}
Displacement S = ut + frac{1}{2} at^2 = 0 (4)+frac{1}{2}times5times4^2
S = 40m.

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

Solution: From the data given in the problem we have,
Initial velocity of the object u = 0,
Acceleration of the object a= 3msec^{-2},
Time of journey t =6sec.
Final velocity of the object v=u+at =0 + 3(6) = 18 m/sec.
Displacement S= ut + frac{1}{2} at^2 = 0 (6)+frac{1}{2}times3times6^2
S= 54m.

3. An object starting from rest moves with uniform acceleration of 4msec^{-2}. Find its displacement i) 5seconds ii) in 5th second iii) 8th second.

Solution : From data given in the problem
Initial velocity of the object u=0,
Acceleration of the object a=4msec^{-2},
1. time t=5 seconds,
Displacement of the object S=ut + frac{1}{2} at^2 = 0 (5)+frac{1}{2}times4times5^2
Displacement of the object in 5seconds S= 50m.
2. Displacement of the body in 5th second =?
Let us substitute n=5 in the formula S_n=u+a(n-1/2)
S_5=0+4(5-1/2) = 4(4.5) =18m.
3. Displacement of the body in 8th second =?
Let us substitute n=8 in the formula S_n=u+a(n-1/2)
S_8=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 1. acceleration 2. time taken for 5m displacement.

solution: 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,
1. acceleration a=?
Substitute the values of u,v and S in the equation V^2-U^2 =2as,
we get (20)^2-(10)^2=2a(5)
300 = 10a or a = 300/10=30msec^{-2}.
2. 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 2.5msec^{-2}and its velocity increases to 75m/sec. Find 1. Initial velocity of the car, 2. Displacement of the car in 10sec, 3. Displacement of the car in first 5sec and last 5sec, what is your inference.

Solution: From the data given in the problem,
Initial velocity of the car u = x (say),
Final velocity = 75m/sec,
Acceleration a = 2.5msec^{-2},
Time of journey t=10sec.
1. 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.
2. Let the displacement of the car in 10sec be S
Substitute the values of u,a and t in the equation S=ut +frac{1}{2} at^2
We get S = 50(10)+frac{1}{2} (2.5)(10)^2 ;
S = 500+125 ; s=675 m
3. Let the displacement in first 5sec be S_1
Substitute t=5 sec and thve values of u and a in the equation S_1=ut +frac{1}{2} at^2
we get S_1=50(5) +frac{1}{2} (2.5)(5)^2
S_1 = 250 + 31.25 = 281.25 m.
Let the displacement in next 5sec be S_2
S_2 = S –S_1
S_2 = 675 – 281.25 = 393.75 m
We can observe that, even though the time of journey is same , S_2>S_1
Displacement of the body in second half S_2 is greater than in the first half time S_1 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?

Solution: 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 v^2-u^2 = 2aS- – – – – (1)
and a = frac{(v-u)}{t} – – – – – (2)
From equations (1) and (2) we get v^2-u^2=2frac{(v-u)}{t}S
simplifying it we get S = frac{(v^2-u^2)}{2}frac{t}{(v-u)} = frac{(v+u)}{2}times t
Substitute the values of u,v and t we get S = frac{(24+0)}{2}times6.7 =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?

Solution:
Method I: Let the total distance between the places be S.
Time taken to cover First half t_1= frac{S}{2times30} = frac{S}{60} hours.
Time taken to cover Second half t_1= frac{S}{2times90} = frac{S}{180} hours.
Total time t = t_1+ t_2 =frac{S}{60}+frac{S}{180}=frac{4S}{180} = frac{S}{45}.
Average speed = Total distance /total time.
= S/t = frac{(S)(45)}{S}
= 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 = frac{2xy}{x+y}.
Average speed = frac{2(30)(90)}{30+90} = 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.

Solution: 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= 40m/sec^2
Substitute the values of u,a and t in the equation S= ut +frac{1}{2} at^2
S=(0)(10)+frac{1}{2} (40)(10)^2 = 0+2000 = 2000m.

9. A body moving with uniform acceleration covers 6m in 2^{nd} second and 16m in 4^{th} second. Calculate the initial velocity, acceleration and distance moved in 6^{ th } second.

Solution: From the data given in the problem
Distance moved in 2^{nd} second s_2 = 6m,
Distance moved in 4^{th} second s_4 = 16m,
s_4 – s_2 = a(4-2) =2a
therefore 2a = s_4 – s_2=16-6 =10
a = 5 m/s^2
Substitute the values of s_2,a and n=2 in the equation s_2 =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 6^{th} second=-1.5+5(5.5)
6^{th} second =-1.5+27.5 =26m.

10. An object started traveling with a velocity 2m/sec moves with an uniform acceleration of 3 m/s^2. 1. Find the ratio of displacements in: 1. 1^{st},3^{rd},5^{th} seconds 2. 2^{nd},4^{th}, and 6^{th} seconds. 2. Find the ratio of velocities: 1. 1^{st},3^{rd},5^{th} seconds 2. 2^{2d},^{th},6^{th} seconds.

Solution: From the data given in the problem
Initial velocity u=2 m/sec,
Acceleration a =3 m/sec^2,
i- a) The ratio of displacements in 1^{st},3^{rd},5^{th} seconds
From the formula s_1:s_3:s_5 = (2u+a) : (2u+5a) : (2u+9a)
s_1:s_3:s_5 = (4+3) : (4+15) : (4+27)
s_1:s_3:s_5 = 7:19:31 .
i-b) The ratio of displacements in 2^{nd},4^{th},6^{th} seconds
From the formula s_2:s_4:s_6 = (2u+3a) : (2u+7a) : (2u+11a)
s_2:s_4:s_6 = (4+9) : (4+21) : (4+33)
s_2:s_4:s_6 = 13 : 25 : 37 .
ii -a) From the formula v_1:v_2:v_3: . . . . . . . . :v_n= (u+a) : (u+2a) : (u+3a) : . . . . . . . . . : (u+na).
The ratio of velocities a) 1^{st},3^{rd},5^{th} seconds
v_1:v_3:v_5 = (u+a) : (u+3a) : (u+5a) = 5 :11 : 17 .
ii- b) The ratio of velocities a) 2^{nd},4^{th},6^{th} seconds
v_2:v_4:v_6 = (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?

Solution: The displacement of the body in first 2 sec S_1 =200cm,
Let the initial velocity = u(say) ,
Acceleration = a(say), time t_1 =2sec
S_1 = ut_1+frac{1}{2} at_1^2,
Substitute the value of t_1 in above equation, we get
S_1 = 2u+frac{1}{2} a(2)^2,
S_1 = 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 S_2 =420 cm,
time t_2=6 sec,
substitute theses values
in the equation S_2 = ut_2+frac{1}{2} at_1^2,
S_2 = 6u+frac{1}{2} a(6)^2=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/sec^2,
Substitute value of a in Eq(1) we get u-15 = 100,
u = 115 cm/sec.
The velocity at the end of 6{th} second v=u+at_2
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 16.5m/sec^2 for 13.1 sec. It runs at constant speed for 69.7s and slows down at a rate of 3.45 m/sec^2until it stops at the next station. What is the total distance covered?

Solution: From the data given in the problem,
Initial velocity of the train u = 0 m/sec,
Acceleration a = 16.5 m/sec^2,
time t = 13.1 sec,
Velocity after 13.1 sec v=?
and the distance traveled S_1 =?
Substitute the values in the equation v =u+at
we get v= 0+(16.5)(13.1) = 216.15 m/sec.
Substitute in equation S_1 = ut + frac{1}{2} at^2
we get S_1 = (0)(13.1)+frac{1}{2} (16.5)(13.1)^2
S_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 S_2 during this time.
v=216.15 m/sec, t = 69.7 sec S_2=?
substitute the values in equation S_2= vt = (216.15)(69.7)=15065.66 m
Finally the trains decelerates at the rate of 3.45 m/sec^2 and comes to rest. Find distanceS_3 traveled before coming to rest.
Acceleration a = -3.45 m/sec^2 ,
Initial velocity u = 216.15m/sec,
Final velocity v=0,
distance traveled S_3 =?
substitute the values in v^2 - u^2 = 2as,
we get 0^2 - 216.15^2 = 2(-3.45) S_3
-6.90 S_3 = -46720.82
S_3 = 6771.13.
Total distance traveled by train S = S_1 +S_2+ S_3 = 1415.78+15065.66+6771.13=23252.57 m or 23.252 km.