**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.

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