www.mysmartschool.com

Equation of motion

(i) Following are of the equations of motion -

v = u + at

v2 = u2 + 2as

(ii) The distance covered in nth second –

(iii) Above all the equations of motion are applicable only when acceleration is constant

(iv) If direction of initial velocity and acceleration be opposite then above equations

v = u - at

v2 = u2 - 2as

(v) If a body moves with uniform acceleration and velocity changes from v1 to v2 in a time interval then –

(vi) If a body start to move from rest and moves with constant acceleration then

(a) The ratio of distances covered in first one sec, two sec, three sec …………………… is equal to 11:22:32: …………….. i.e. 1:4:9:………

(b) The ratio of distances covered in 1st, 2nd, 3rd ………………….. is 1 : 3 : 5 : ………….

(c) The ratio of  velocities after 1sec, 2 sec, 3 sec …………………. is 1 : 2 : 3 : …………..

(v) A body is thrown vertically upwards from the top A of tower. It reaches the ground in t1 seconds. If it is thrown vertically downwards from A with the same speed, it reaches the ground in t2 sec. If it is allowed to fall freely from A, then the time it takes to reach the ground t is given by.

Q1.       A man standing on the edge of a cliff throws a stone straight up with initial speed u and then throws another stone straight down with the same initial speed and from the same position. Find the ratio of the speed the stones would have attained when they hit the ground at the base of the cliff

Ans.     As the stone thrown vertically up will come back to the point of projection with the same speed, Both the stones will moves downward with same initial velocity, so both will hit the ground with same speed   -v2 = u2 + 2gh

Hence, the ratio of speeds stained when they hit the ground is 1 : 1

Notes: However the stone projected up will take (2u/g) time more to reach the ground than the stone projected downwards.

Q2.       A juggler throws balls into air.  He throws one whenever the previous one is at its highest point. How high do the balls rise if he throws n balls each second? Acceleration due to gravity is g.

Ans.     As the higher is throwing n balls each second and 2nd when the first is at its highest point, so the time taken by one ball to reach the highest, point.

and as at highest point  v = 0, from 1st  equation of motion –

Now from 3rd equation of motion i.e.

Q3.       A pebble is thrown vertically upwards from a bridge with an initial velocity of 4.9 m/s. it strikes the water after 2sec. If acceleration due to gravity is 9.8 m/s2 (a) what is the highest of the bridge (b) with what velocity does the pebble strikes the water.

Ans.     Taking the point of projection as origin and downward direction as positive.

(a) By 2nd equation of motion i.e.

h = 9.8 m

(u is taken to be negative as it is upward)

(b) By 1st equation of motion i.e.

v = u + at

v = -4.9 + 9.8 x 2

v = 14.7 m/s

Q4.       A rocket is fired vertically up from the ground with a resultant vertical acceleration of 10 m/s2. The fuel is finished in 1 minute and it continues to move up.

(a)     What is the maximum height reached

(b)    After how much time from then will the maximum height be reached?

(takes g= 10 m/s2

Ans.     The distance traveled by the rocket during burning interval (1 min = 60 sec.) in which resultant acceleration is vertically upwards and d10 m/s2 will be-

and velocity acquired by it will be-

v = ut + at

= 0 + 10 x 60 = 600 m/s                         (2)

Now after 1 minute the rocket moves vertically up with initial velocity of 600 m/s and acceleration due to gravity opposes its motion, so it will go to a height h2 till its velocity becomes zero such that –

v2 = u2 – 2gh2

0 = (600)2 – 2 x 10 x h2

or h2 = 18000m                          { because g = 10m /s2}

Now from 1 and 3, the maximum height reached by the rocket from the ground –

H = h1 + h2 = 18000 + 18000

= 36000 m

= 36 km

(b) As after burning of fuel the initial velocity from (2) is 600 m/s and gravity opposes the motion of rocket, so from 1 st equation of motion time taken by it to reach the  maximum height (for which v = 0)

v = u – gt

0 = 600 – 10t

t = 60 sec

Q5.       A body is released from a height and falls freely towards the earth, exactly 1 sec later another body is released. What is the distance between the two bodies 2 sec the release of the second body. If g= 9.8 m/s2

Ans.     According to the problem, 2nd body falls for 2 sec, So that

while, 1st has fallen for 2 + 1 = 3 sec, So

Hence Separation between two bodies 2 sec after the release of 2nd body,

d = 24.5 m

Q6.       A block slides down a smooth inclined lane when release from the top, while another falls freely from the same point. Which one of them will strike the ground?

(i) Earlier                       (ii) with greater speed

Ans.     Acceleration of block along the inline place be “g Sin q”. If time taken by body (the body moving along the incline plane) to fall h height vertically be ts then –

while in case of Free fall.

Hence, falling body reaches the ground first.

Both reach the ground with same speed. (not same velocity, as for falling body direction is vertical while for sliding body along the plane downwards.

Q7.       The acceleration of a particle moving in a straight line varies with its displacement as –

a = 2s

velocity of the particle is zero at zero displacement – Find the corresponding velocity – displacement equation.

Ans.     According to the problem –

a= 2s

v. dv = 2sds

v2 = 2s2

Q8.       The retardation of a particle moving in a straight line is proportional to its displacement (proportionality constant being unity). Find the total distance covered by the particle till it come to rest. Given that velocity of particle is v0 at zero displacement.

aa - s

a = -ks

or a = -s            {because K = l}

v.dv = -s ds

or  s = v0

Q9.       A particle moves in a straight line with a uniform acceleration a. Initial velocity of particle is zero. Find the average velocity of the particle in first “s” distance.

Ans.     From,

Hence,

Average velocity = s/t

Q10.     A body is moved along a straight line by a machine delivering constant Power. Find the relation between distance traveled and time.

Ans.     The work done on the body converts in the kinetic energy of the body.

Q11.     A ball is dropped from the top of a tower 100 m high. Simultaneously another ball is thrown upward with a speed of 50m/s. After what time do they cross each other?

Ans.     Let balls A and B cross each other at point p and at high h and take positive direction upward.

Now for A –

(100- h) = 0 – ½ gt2 g be negative be cause or h = 100 – ½ gt2 …. (1) its direction be downward

For B –

h = 50t – ½ gt2   …… (2)

From 1 and 2

or 100 = 50t

or t= 2 sec

Q12.     A truck and a car are bought to a halt by application of same braking force. Which one will come to stop in a shorter distance if they are moving with same (a) velocity (b) kinetic energy (c) momentum?

Ans.     On applying brakes the body is brought to rest, so V = 0 and. a = [-F/m] If S is the distance traveled in stopping then from 3 rd equation of motion –

v2 = u2 + 2aS

(a) If u is same then from (1)

S a m

Since, mass of car is less than that of truck, So car will stop is shorter distance.

(b) In terms of kinetic energy equation (1) written as  -

Since, kinetic energy and breaking force be same. Hence both will stop after traveling same distance.

(c) In terms of momentum above equations (1) written as -

If P is same then –

Since, mass of truck is more than that of car. So truck will stop in a shorter distance.

Q13.     A passenger is standing “d” m away from a bus begins to move with constant acceleration a. To catch the bus, the passenger runs at a constant speed v towards the bus. What must be the minimum speed of the passenger so that he may catch the bus?

Ans.     Let passenger catch the bus after time t, and at paint P.

From the second equation of motion distance traveled by the bus

S1 = 0 + ½ at2 –-- (1)

and in time t distance traveled by passenger S2 = ut + 0 –(2)

Now the passenger catch the bus if –

d + S1 = S2

d + ½ at2 = ut

½ at2ut + d = 0

So, the passenger will catch the bus if t is real i.e.

Exercise - 4

Q1.       A body is dropped from the top of the tower and falls freely. The distance covered by it after n sec is directly proportional to –

(i) n2

(ii) n

(iii) 2n – 1

(iv) 2n2 – 1

Ans.     (i)

Q2.       A particle experiences constant acceleration for 20 sec after starting from rest. If it travels a distance S1 in the first 10 sec and distance S2 in the next 10 sec then –

(i) S2 = S1

(ii) S2 = 2S1

(iii) S2 = 3S1­

(iv) S2 = 4S1

Ans.     (iii)

Q3.       The velocity acquired by a body moving with uniform acceleration is 30 m/s in 2 sec and 60 m/s in 4 sec, the initial velocity is

(i) zero

(ii) 2m/s

(iii) 4m/s

(iv) 10m/s

Ans.     (i) zero

Q4.       The splash is heard after 2.05 sec after the stone is dropped in to well of depth 19.6m. The velocity of sound is –

(i) 342 m/s

(ii) 372 m/s

(iii) 392 m/s

(iv) 352 m/s

Ans.     (iii)

Q5.       When the speed of car is v, the minimum distance over which it can be stopped is s. If the speed becomes nv, what will be the minimum distance over which it can be stopped during same time –

(i) s/n

(ii) ns

(iii) s/n2

(iv) n2s

Ans.     (iv)

Q6.       A point moves with uniform acceleration and V1, V2 and V3 denote the average velocities in the three successive intervals of time t1, t2 and t3 which of the following relations is correct -

(i) (v1 – v2) : (v2 – v3) = (t1 – t2) : (t2 + t3)

(ii) (v1 – v2) : (v2 – v3) = (t1 + t2) : (t2 + t3)

(iii) (v1 – v2): (v2 – v3) = (t1 – t2) : (t1 + t3)

(iv) (v1 – v2) : (v2 – v3) = (t1 – t2) : (t2 – t3)

Ans.     (ii)

Q7.       A body is released from the top of a tower of height H metre. After 2sex it is stopped and then instantaneously released. What will be its height after next 2 seconds

(i) (H -5) metre

(ii) (H – 10) metre

(iii) (H – 20) metre

(iv) H – 40) metre

Ans.     (iv)

Q8.       The displacement of a particle after time t is given by –

Ans.     (ii)

Q9.       A stone is dropped from the top of a tower of height h. After 1 sec another stone is dropped from the balcony 20 below the top. Both reach the bottom simultaneously. What is the value of h        (take g = 10m/s2)

(i) 3125 m/s

(ii) 312.5 m

(iii) 31.25 m

(iv) 250.31m

Ans.     (iii)

Q10.     A car starts from rest and moves with constant acceleration. The ratio of distance covered in the nth second to that covered in n second is –

Ans.     (iii)

Q11.     A person is trowing balls in to the air one after the other. He throws the second ball when first ball is at the highest point. If he is throwing two balls every second, how high do they rise -

(i) 5m

(ii) 3.75 m

(iiii) 2.5m

(iv) 1.25m

Ans.     (iv)

Q12.     The speed of a body is doubled when it moves over a distance of 10m. If the initial speed be u, what will be the speed after further coverage of distance 10m-

(i) Ö 5u

(ii) Ö6u

(iii) Ö7u

(iv) Ö8u

Ans.

Q13.     A particle moves for 8 seconds. It first accelerates from test and then retards to rest. If the retardation be 3 times the acceleration, then time for which it accelerates will be –

(i) 2 sec

(ii) 3 sec

(iii) 4 sec

(iv) 6 sec

Ans.     (iv)

Q14.     A body dropped from the top of the tower covers a distance 7x in the last second of its journey, where x is the distance covered in first seconds, How much times does it take to reach the ground.

(i) 3 sec

(ii) 4 sec

(iii) 5 sec

(iv) 6 sec

Ans.     (ii)

Q15.     A body travels 200 cm in the first two seconds and 220 cm in the next 4 sec with deceleration. The end of 7th second –

(i) 5 cm/s

(ii) 10 cm/s

(iii) 15 cm/s

(iv) 20 cm/s

Ans.     (ii)

Q16.     A body sliding on a smooth inclined plane requires 4 sec to reach the bottom starting from rest at the top. How much time does it take to cover one- further the distance starting from rest at the top –

(i) 1 sec

(ii) 2 sec

(iii) 4 sec

(iv) 16 sec

Ans.     (ii)

Q17.     A ball is thrown vertically upwards with a speed of 10m/s from the top of a tower 200m high and another is thrown vertically downwards with the same speed simultaneously. The time difference between them in reaching the ground in second (g = 10m/s2) is.

(i) 12

(ii) 6

(iii) 2

(iv) 1

Ans.     (iii)

Q18.     An elevator in which a man is standing is moving upwards with a speed of 10m/ s. If the man drops a coin from a height of 2.45 metre, it reaches the floor of the elevator after a time (g = 9.8 m/s2)

Ans.     (ii)

Q19.     One body is dropped while a second body is thrown downwards with an initial velocity of 2m/s simultaneously. The separation between them is 18 metres after a time –

(i) 9 sec

(ii) 4.5 sec

(iii) 18 sec

(iv) 9.8 sec

Ans.     (i)

Q20.     Two car A and B are traveling in the same direction with velocities VA and VB (VA>VB). When the car A is at a distance S behind the car B, the driver of the car A the applies the brakes producing a uniform retardation a, there will be no collision when –

Ans.     (iii)

Q21.     A ball is thrown vertically upwards from the ground G with a speed u. It reaches a point G with a speed u. It reaches a point B at a height h (lower then the maximum height) after time t1.It returns to the ground after time t2 from the instant it was at B during the upward journey. Then t1t2 is equal to

(i) 2h/g

(ii) h/g

(iii) h/2g

(iv) h/4g

Ans.     (iv)

Q22.     A body falling freely from a given height H hits on an inclined plane in its path at a height h. As a result of this impact, the direction of velocity becomes horizontal for what value of h/H, the body will take maximum time to reach the ground –

(i) 3/4

(ii) ½

(iii) ¼

(iv) 2/3

Ans.     (ii)

Q23.     A police party is chasing dacoit in a jeep which is moving at a constant speed v. The dacoit is on a motor cycle. When he is at a distance x from the jeep he accelerates at a constant rate a. Which of the following relations is true, if the police is able to catch the dacoit.

(i) v2 £ 2a x

(ii) v2 £ a x

(iii) v2 ³ 2a x

(iv) v2 ³ a x

Ans.     (iii)

 www.mysmartschool.com