DC pandey

3

Motion in One Dimension Introductory Exercise 3.1

1. Suppose

a particle is moving with constant velocity v (along the axis of x). Displacement of particle in time t1 = vt1 Displacement of particle in time t2 = vt2 ∴ Displacement of the particle in the time interval ∆t ( = t2 − t1) = vt2 − vt1 = v ( t2 − t1) ∴ Average velocity in the time interval v ( t2 − t1) ∆t ′ = ( t2 − t1) =v Now, as the particle is moving with constant velocity (i.e., with constant speed in a given direction) its velocity and speed at any instant will obviously be v. Ans. True.

2. As the stone would be free to acceleration

under earth’s gravity it acceleration will be g. 3. A second hand takes 1 min i.e., 60s to

complete one rotation (i.e., rotation by an angle of 2π rad). ∴ Angular speed of second hand 2π rad = 60 s π = rad s −1 30 Linear speed of its tip = radius × angular speed

= 2.0 cm ×

π rad s −1 30

π cms −1 15 As the tip would be moving with constant speed. π Average speed = cms −1 15 In 15 s the second hand would rotate through 90° i.e., the displacement of its tip will be r 2. ∴ Modulus of average velocity of the tip of second hand in 15 s. r 2 = 15 2 2 = cms −1 15 =

4. (a) Yes. By changing direction of motion,

there will be change in velocity and so acceleration. (b) (i) No. In curved path there will always be acceleration. (As explained in the previous answer no. 3) (ii) Yes. In projectile motion the path of the particle is a curved one while acceleration of the particle remains constant. (iii) Yes. In curved path the acceleration will always be there. Even if the path is circular with constant speed the direction of the acceleration of the particle would every time be changing.

Motion in One Dimension Circumference Speed 2π × 4 cm = = 8π = 25.13 s 1 cm / s

4v 2 2πr v 2 2 v2 = π r 2 2 (1)2 = π 4

5. (a) Time speed =

(b) As particle is moving with constant speed of 1 cm/s, its average speed in any time interval will be 1 cm/s. r 2 |Average velocity|= T/4 =

4r 2 2 2 speed = π  2πr     speed 

| 15

=

= 0.23 cm s −2 6. Distance = Speed × time

D1 = v1t1 D2 = v2 t2

2 2 cms −1 π = 0.9 cms −1 v 2 |Average acceleration|= T/4

Average speed =

=

=

D1 + D2 v1t1 + v2 t2 = t1 + t2 t1 + t2 ( 4 × 2) + (6 × 3) 2+3

= 5.2 ms −1

(where v = speed)

Introductory Exercise 3.2 1. Acceleration (due to gravity).

1 1 = ut + at2  − u ( t − 1) + a( t − 1)2      2 2   st = u + at –

1 a 2

Displacement Velocity

Therefore, the given dimensionally incorrect.

Acceleration

equation

v = t 3/ 4 ds = t 3/ 4 dt

4.

1 2. st = u + at − a is physically correct as it 2 gives the displacement of the particle in tth second (or any time unit). st = Displacement in t seconds − displacement in ( t − 1) seconds

∴

s=

∫t

(given) …(i) 3/ 4

dt =

3 +1 t4

3 +1 4

+c

4 7 /4 t +c 7 i.e., s ∝ t7 /4 Differentiating Eq. (i) w.r.t. time t, 3 d2 s 3 4 − 1 = t dt2 4 ⇒ a ∝ t −1/ 4 or

s=

5. Displacement (s) of the particle

is

3. Yes. When a particle executing simple

harmonic motion returns from maximum amplitude position to its mean position the value of its acceleration decreases while speed increases.

s = ( 40 × 6) +

1 ( − 10) 62 2

= 240 − 180 = 60 m (in the upward direction) Distance covered ( D) by the particle Time to attain maximum height

16 | Mechanics-1 40 = 4s < 6 s 10 It implies that particle has come back after attaining maximum height (h) given by u2 h= 2g = ∴

( 40)2 = 80 m 2 × 10

D = 80 + (80 − 60) = 100 m

v = 40 − 10t dx ∴ = 40 − 10t dt or dx = ( 40 − 10t) dt or x = ∫ ( 40 − 10t) dt or x = 40t − 5t2 + c As at t = 0 the value of x is zero. c=0 ∴ x = 40t − 5t2 For x to be 60 m. 60 = 40t − 5t2 or t2 − 8 t + 12 = 0 ∴ t = 2 s or 6 s Displacement in time t 7. Average velocity = t 1 2 ut + at 2 = t 1 = u + at 2 6.

8.

∴

v2 = v1 + at at = v2 − v1 Displacement in time t t 1 2 v1t + at 2 = t 1 = v1 + at 2 v2 − v1 = v1 + 2 v1 + v2 = 2

Ans. True.

∴

=

125 = 0 ⋅ t +

9.

1 2 gt 2

t = 25 s 125 m Average velocity = 5s

(downwards)

= 25 m/s

(downwards)

⇒

2

…(i

v = 10 + 5t − t

10.

) dv = 5 − 2t dt At t =2s a=5−2×2 = 1 m/s2 From Eq. (i), dx = 10 + 5t − t2 dt ∴ x = ∫ (10 + 5t − t2 ) dt 5t2 t 3 or x = 10t + − +c 2 3 As, at t = 0 the value of x is zero c=0 5 t3 ∴ x = 10t + t2 − 2 3 Thus, at t = 3 s 5 33 x = (10 × 3) + (3)2 − 2 3 = 30 + 22.5 − 9 = 43.5 m a=

∴

→

^

11. u = 2 i m/s →

a = (2 cos 60° ^i + 2 sin 60° ^j ) m/s2 = (1 ^i + 3 ^j) m/s2 y

Average velocity =

a = 2 m/s2 60° u = 2 m/s →

→

→

v=u+at = 2 ^i + (1 ^i + 3 ^j) 2 = 4 ^i + 2 3 ^j

x

Motion in One Dimension →

Part II Yes. As explained below.

|v|= 42 + 12 = 2 7 m/s → → 1→ s = u t + a t2 2 1 ^ = (2 i) 2 + (1 ^i + 3 ^j) 22 2

→

v = 2 ^i + 2t ^j implies that initial velocity of

the particle is 2 ^i m/s2 and the acceleration is 2$j m/s2 → 1 ∴ s (at t = 1 s) = (2 ^i × 1) + (2 ^j) 12 2

= 4 ^i + 2 ^i + 2 3 ^j

= (2 ^i + ^j) m

= 6 ^i + 2 3 ^j

13. x = 2t and y = t2

→

|s |= 36 + 12

∴

→

v = (2 ^i + 2t ^j)m/s

…(i)

→

dv = 2 ^j dt →

a = 2 ^j m/s2

From Eq. (i), →

ds = (2 ^i + 2t ^j ) dt s = ∫ (2 ^i + 2t ^j ) dt

2

or, x2 = 4 y (The above is the equation to trajectory) x = 2t → dx ∴ = 2 i.e., v x = 2 ^i dt y = t2 → dy ∴ = 2t i.e., v y + 2t ^j dt →

Thus,

→

∴

x y =    2

∴

=4 3m 12. Part I

| 17

→

→

v = vx + v y = (2 ^i + 2t ^j) m/s

→

s = 2t ^i + t2 ^j + c

→

Taking initial displacement to be zero. →

s (at t = 1 s) = (2 ^i + ^j) m

→

a=

dv = 2 ^j m/s2 dt

Introductory Exercise 3.3 1. At t = t1 v

s

α

t2 β

β

t1 0

θ t1

t2

φ

v = tan θ As θ < 90°, vt1 is + ive. At t = t2 vt 2 = tan φ As φ > 90°, vt 2 is − ive. Corresponding v-t graph will be

t

t

Acceleration at t = t1 : at1 = tan α As α < 90°, a t1 is + ive constant. Acceleration at t = t2 at 2 = tan β As β < 90°, at 2 is + ive constant. 2. Let the particle strike ground at time t

velocity of particle when it touches ground

18 | Mechanics-1 1 mg2 t2 2 i.e., KE ∝ t2 . While going up the velocity will get − ive but the KE will remain. KE will reduce to zero at time 2 t when the particle reaches its initial position.

Corresponding velocity-time will be

would be gt. KE of particle will be

Velocity (m/s)

4 KE

4.

t =

KE =

2t

2h g

time

8

t (s)

h 2 = tan θ = ( t − 2) (2 − 1) h = 2 ( t − 2)

⇒ 2

a (m/s )

1 1 2h mg2 t2 = mg2 2 2 g

= mgh

2

(t–2)

θ

3. Speed of ball (just before making first

collision with floor) = 2gh = 2 × 10 × 80 = 40 m/s Time taken to reach ground 2 × 80 2h = = =4s g 10 Speed of ball (just after first collision with floor) 40 = = 20 m/s 2 Time to attain maximum height −20 t= = 2s −10 ∴ Time for the return journey to floor = 2 s. Speed (m/s)

1

2 θ

t h

Particle will attain its initial velocity i.e., net increase in velocity of the particle will be zero when, area under a-t graph = 0 (1 + 2) × 2 ( − h) ( t − 2) + =0 2 2 or 3 − ( t − 2)2 = 0 or ( t − 2)2 = 3 or t−2= ± 3 or t=2± 3 Ans . At time t = 2 + 3 s (t = 2 − 3 not possible).

4

8

t (s)

Introductory Exercise 3.4 1. Relative acceleration of A w.r.t. B

a AB = ( + g) − ( + g) = 0 2. Velocity of A w.r.t. B = v A − vB

∴ Relative displacement (i.e., distance between A and B) would be 1 s = (v A − vB ) t + a AB t2 2

Motion in One Dimension or

sin 30° sin θ sin (180° − 30° + θ) = = 150 t 20t 500 × 103 1 sin θ = 300 20 1 θ = sin −1    15 15 sin (30° + θ) 1 Now = 3 300 t 500 × 10 224 5000 or = sin 30° cos θ + cos 30° sin θ 3t

s = (v A − vB ) t tan θ = (vA–vB)

s

θ

θ

0

t

3. In figure, u = speed of boat

v = speed of river flow B

C Actual path of boat

Boat sailing direction

or

u d = 400 m

⇒

A v

Time to cross river AB  BC AC  = = =  2 u  v u + v2   400 m = 10 m/s = 40 s v BC = AB u 2 m/s = × ( 400 m ) 10 m/s = 80 m 4. Let C be the point along which pilot

should head the plane. B

30°

5 00

km

North

θ

Drifting due to wind

Speed = 150 m/s

A Wind/speed = 20 m/s

Apply sine formula in ∆ ABC

| 19

East

1

5000 1 224 3 1 = + ⋅ 3t 2 15 2 15 5000 = 0.5577 3t 5000 t= 3 × 0.5577 = 2989 s = 50 min

5.

10 m A

B

a A = 1 m/s2 , aB = 2 m/s2 v A = 3 m/s, vB = 1 m/s Acceleration of A w.r.t. B = 1 − 2 = − 1 m/s2 Velocity of A w.r.t. B = 3 − 1 = 2 m/s Initial displacement of A w.r.t. B = − 10 m At time relative displacement of A w.r.t. B 1 s = − 10 + 2 t + ( −1) t2 2 or s = − 10 + 2t − 0.5t2 For s to be minimum ds =0 dt or 2 − (0.5 × 2t) = 0 i.e., t = 2s ∴ smin = − 10 + (2 × 2) − 0.5 × (2)2 = − 10 + 4 − 2 = −8m Minimum distance between A and B = 8 m.

20 | Mechanics-1 AIEEE Corner Subjective Questions (Level 1) 1. (a) D = v1t1 + v2 t2

km km 1  = 60 × 1 h + 80 × h    h h 2  = 100 km (b) Average speed 100 km = = 66.67 km/h. 1.5 h 2. (a) Displacement in first two seconds

= ( 4) (2) +

1 (6) 22 2

= 20 m

20 m ∴ Average velocity = = 10 m/s 2s (b) Displacement in first four seconds 1 = ( 4) ( 4) + (6) 42 = 64 m 2 ∴ Displacement in the time interval t = 2 s to t = 4 s = 64 − 20 = 44 m 44 m ∴ Average velocity = = 22 m/s. 2s

(b) Average velocity = =

Net displacement Time taken

− 60 m 6.1 s

= − 9.8 m/s (downwards) = 9.8 m/s Σ vt 4. Average velocity = Σt vt0 + 2 vt0 + 3vT 2.5v = t0 + t0 + T ∴ or

5t0 + 2.5T = 3 t0 + 3 T T = 4 t0

5. (a) Average acceleration

Final velocity − initial velocity ( 4 + 8) s 0−0 = 12 = 0 m/s2 v (c) a = tan θ = max 4 =

v vmax

3. Let the particle takes t time to reach

ground. +20 m/s

θ

h

4s

– 60 m

∴

− 64 = 20 t +

5t2 − 20t − 64 = 0 t = 6.1 s If the particle goes h meter above tower before coming down 0 = (20)2 + 2 ( −10) h ⇒ h = 20 m Total distance moved (a) Average speed = Time taken (20 + 20 + 60) = = 16.4 m/s 6.1 i.e.,

12s time

vmax (Q a = 4 m/s2 ) 4 i.e., vmax = 16 m/s Displacement of particle in 12 seconds = Area under v-t graph v = 12 × max 2 12 × 16 = 2 = 96 m Average velocity Displacement (96 m) at 12 s = Time (12 s)

∴ 1 ( − 10) t2 2

8s

4=

= + 8 m/s

Motion in One Dimension (b) As the particle did not return back distance travelled in 12 s = Displacement at 12 s ∴ Average speed = 8 m/s. 21 6. (a) Radius ( R) of circle = m 22 ∴ Circumference of circle = 2πR 22 21 m =2× × 7 22 = 6m Speed (v) of particle = 1 m/s ∴ Distance moved by particle in 2 s = 2 m Thus, angle through which the particle moved 2 2π = × 2π = = 120° 6 3 Magnitude of Average velocity Magnitude of displacement = Time ( = 2 s)

7. Position vector at t = 0 s →

r1 = (1 ^i + 2 ^j) m Position vector at t = 4 s →

r2 = (6 ^i + 4 ^j) m (a) Displacement from t = 0 s to t = 4 s →

= (6 ^i + 4 ^j) − (1 ^i + 2 ^j) = (5 ^i + 2 ^j) m ∴ Average velocity =

(b) Average acceleration Final velocity − Initial velocity = 4s

120° O

AB 2R sin 60° = 2 2 3 =R 2 21 3 21 3 m/s = = 22 2 44 (b) Magnitude of average acceleration → → vB − v A  =    2s   120° 2 v sin 2 = 2 =

[Q

→

→

|v A |= |v B |= v = 1 m/s) = v sin 60° 3 m/s2 = 2

(5 ^i + 2 ^j) m 4s

= (125 . ^i + 05 . ^j) m/s

=

(2 ^i + 10 ^j) − ( 4 ^i + 6 ^j) 4

=

−2 ^i + 4 ^j 4

A(t = 0 s)

(t = 2s) → vB

→

= ( r 2 − r1 )

→ vA

B

| 21

= ( −05 . ^i + ^j) m/s2 (c) We cannot find the average speed as the actual path followed by the particle is not known. Uniform acceleration (a) One dimensional motion 8. If at time t the vertical displacement

between A and B is 10 m 1 2 1 gt − g( t − 1)2 = 10 2 2 or or or

t2 − ( t − 1)2 = 2 t2 − ( t2 − 2t + 1) = 2 2t = 3 t = 1.5 s

9. The two bodies will meet if

Displacement of Displacement of first after attaining = second before attaining highest highest point point

22 | Mechanics-1 v0 t − or or

or

10.

⇒ T O

1 2 1 gt = v0 ( t − t0 ) − g( t − t0 )2 2 2 1 2 0 = − v0 t0 − g( t0 − 2t0 t) 2 1 2 gt0 t = gt0 + v0 t0 2 1 gt0 + v0 t= 2 g t0 v0 = + 2 g 1 5 = gt2 2 t = 1s 5m A 25 m

H W

∴ or or t=

2t0 ± ( −2t0 )2 − 41( − t20 )

2 2t0 ± 2t0 2 = 2 = t0 + t0 2 (− absurd) = (1 + 2) t0 = 2141 . t0

Time interval 1s

ive

t second for A

12. (t–1) second for B

⇒

v=0 A

B

5 m/s H

For A H = 0. t + H=

1 2 gt 2

1 2 gt 2

15m

…(i)

1 …(ii) g( t − 1)2 2 1 2 1 or gt − g( t − 1)2 = 25 2 2 [Substituting value of H from Eq. (i)] 1 g[ t2 − ( t − 1)2 ] = 25 2 t2 − ( t2 − 2t + 1) = 5 2t − 1 = 5 ⇒ t=3s Substituting t = 3 s in Eq. (i) 1 H = × 10 × 32 = 45 m 2 1 2 11. s = at0 Forward motion 2 H − 25 =

Backward motion

u

(a) For H 02 = u2 + 2( −10) H i.e., 20H = 325 or H = 1625 . m

For B

v = at0

being

52 = u2 + 2 ( −10)15 u2 = 325

R

or

sign

From the begining of the motion the point mass will return to the initial position after time 3.141t0 .

B u=0

E

1 ( − a) t2 2 1 1 − at20 = ( at0 ) t − at2 2 2 2 2 − t0 = 2t0 t − t 2 t − 2t0 t − t20 = 0 − s = ( at0 ) t +

(b) For t 0 = 325 + ( −10) t 325 t= 10 = 1. 8 s

∴

13.

At rest

a x

u

a

15 m/s

60 m 6.0 s

(a) …(i) 152 = u2 + 2a × 60 and …(ii) 15 = u + a × 6 Substituting the value of 6a from Eq. (ii) in Eq. (i) 225 = u2 + 20 (15 − u)

Motion in One Dimension 2

i.e.,

u − 20 u + 75 = 0 (u − 15)(u − 5) = 0 ∴ u = 5 m/s (15 m/s being not possible) (b) Using Eq. (ii) 5 a = m / s2 3 (c) u2 = 02 + 2ax u2 (5)2 i.e., x= = 2a 2 × 5 3 = 7.5 m 1 (d) s = at2 2 1 5 = × × t2 2 3 5 = t2 6 t(s )

0

1

3

v (m/s)

0

5/6

7.5

6

Starts A

…(iii) 9

12

30 15 t (s)

12

Differentiating Eq. (iii) w.r.t. time t 5 v= t 3 t(s )

0

3

6

9

12

v (m/s)

0

5

10

15

20

v (m/s) 20 15 10 5 6

9

12

Stops B B

s2

Journey A to P vmax = 0 + xt1 and v2max = 2xs1 v ⇒ t1 = max x

60

3

t2

4 min

30 67.5 120

9

t1 4 km

90

6

a = –y

s1

120 s (m)

3

a=+x

14.

| 23

t (s)

…(i) …(ii)

Journey P to B 0 = vmax + ( − y) t2 2 and vmax = 2 ys2 v ⇒ t2 = max y vmax vmax ∴ + = t1 + t2 = 4 x y 1 1  vmax  +  = 4  x y From Eq. (ii) and Eq. (iv) v2 v2 s1 + s2 = max + max 2x 2y v2max  1 1  or 4= + 2  x y  or

…(iii) …(iv)

…(v)

…(vi)

Dividing Eq. (vi) by Eq. (v) vmax = 2 Substituting the value of vmax in Eq. (v) 1 1 (Proved) + =2 x y 15. Let acceleration of the particle be a using 2m 4 0 t = 10 s t = 6 s v=0

t=0s

+ive x-axis

v = u + at 0 = u + a6 u ∴ a=− 6 (a) At t = 10 s, s = − 2 m 1 −2 = u × 10 + a × 102 2 or −2 = ( −6a) 10 + 50a or −10a = − 2 or a = 0.2 m/s 2

24 | Mechanics-1 →

(b) v(at t = 10 s) = u + a 10 = − 6a + 10a = 4a = 08 . m/s

→

u = 0 m/s

→

a 2 = − 4 ^j m/s2 (t = 2 s to t = 4 s) t2 = 2 s (a) Velocity →

→

→

→

a=

16.

→

= 0 + (2 ^i) 2

F 10 N north = m 2 kg

= 4 ^i

2

= 5 m/s , north ^

= 5 j m/s

→

→

→

v2 = (8 ^i − 8 ^j ) m/s

or

^

= 10 i m/s

(b) Co-ordinate of particle → → → 1 s 1 = s 0 + u t1 + → a 1 t12 2

→

using v = u + a t →

v = 10 ^i + (5 ^j × 2) ^

→

= 4 ^i + (2 ^i − 4 ^j ) 2

u = 10 m/s, east

→

→

v2 = v1 + (a 1 + a 2 )t2

2

→

→

→

v1 = u + a 1 t1

(c) Two or three dimensional motion

= (2 ^i + 4 ^j) + (0)(2) +

^

= 10 i + 10 j

1 ^ 2 (2 i) 2 2

= 2 ^i + 4 ^j + 4 ^i

→

|v|= 10 2 m/s

= 6 ^i + 4 ^j

→

v = 10 2, north-east → → 1→ using s = u t + a t2 2 1 ^ = (10 i × 2) + (5 ^j)22 2

1 ( a1 + a2 ) t22 2 1 = 6 ^i + 4 ^j + ( 4 ^i) 2 + (2 ^i − 4 ^j) 22 2

→

→

→

s 2 = s 1 + v1 t2 +

= 6 ^i + 4 ^j + 8 ^i + 4 ^i − 8 ^j

= (20 ^i + 10 ^j) m

= 18 ^i − 4 ^j

→

|s|= 202 + 102

Co-ordinate of the particle [18 m , − 4 m ]

= 10 5 m 20 cotθ = =2 10 θ = cot−1 2

→

^

→

^

→

18. u = (2 i − 4 j) m/s, s 0 = 0 m →

a = ( 4 ^i + ^j) m/s2

North

(a) Velocity →

θ East

→

s = 10 5 m at cot−1(2) from east to north.

→

^

^

17. s 0 = (2 i + 4 j) m →

→

= (2 ^i − 4 ^j) + ( 4 ^i + ^j)2

^j O ^ i

→

v=u+at

→ s

a 1 = 2 ^i m/s2 (t = 0 s to t = 2 s) t1 = 2 s

= (10 ^i − 2 ^j ) m/s (b) Co-ordinates of the particle → → → 1→ s = s 0 + u t + a t2 2 → 1 ^ = 0 + (2 i − 4 ^j) + ( 4 ^i + $j )22 2

Motion in One Dimension = 10 ^i − 2 ^j

=

∴ Co-ordinates of particle would be [10 m , − 2 m ] u = 8 ^j m/s

→

|a av|= 52 + 202

→

a = ( 4 ^i + 2 ^j) m/s2

→

= 206 . m/s 20 tan θ = 5 i.e., θ = tan1( 4) Non uniform acceleration.

→

s0 = 0 →

s = 29 ^i + n ^j

→

→

→

(a) s = s 0 + u t +

1→ 2 at 2

21. x = 2 + t2 + 2t 3

1 ^ ( 4 i + 2 ^j) t2 2

→

29 ^i + n ^j = 0 + (8 ^j) t +

(a) At t = 0, x = 2 m (b)

29 ^i + n ^j = 8 ^j t + 2 ^i t2 + ^j t2 Comparing the coefficients of ^i and ^j 29 = 2t2 n = 8 t + t2 29 t= = 3.807 s 2

and ⇒

Substituting value of t in Eq. (ii) n = 8 × 3.807 + (3.807)2 = 44.95 29 (b) Speed at t = s 2 →

→

…(i) …(ii)

d2 x = 2 + 12t dt2 ( a) t = 2 s = 2 + (12 × 2) = 26 m/s2 dv 22. a= −v dx = − (10) × 3 = − 30 m/s2 23. s = t 3 − 9 t2 − 15t

ds = 3 t2 − 18 t − 15 dt dv ie, a= = 6t − 18 dt Acceleration (a) in the interval 0 ≤ t ≤ 10 s will be maximum at t = 10 s a(at t = 10 s) = (6 × 10) − 18 = 42 m/s2

∴

→

= 8 ^j + ( 4 ^i + 2 ^j)(3.807) = (2 × 3.807) ^i + {( 4 × 3.807) + 8 } ^j = (7.614 ^i + 23.228 ^j) m/s →

speed = |v|= (7.614)2 + (23.228)2 = 24.44 m / s 20. s 0 = 5 i m →

→

Average velocity v av ^

=

^

→

→

s − s0 = 002 . ^

(51 . i + 0.4 j) − (5 i) 002 .

v=

a = 3 − 2t dv = 3 − 2t dt

24.

^

at t = 002 . s, s = (51 . ^i + 0.4 ^j) m

dx = 2t + 6t2 dt  dx  = 0 m/s    dt  t = 0 s

(c)

v=u+at

→

01 . ^i + 0.4 ^j 002 .

= 5 ^i + 20 ^j m/s

→

19.

| 25

or

∫ dv = ∫ (3 − 2t) dt

or

v = 3 t − t2 + c

or

v = 3 t − t2 + v0 [as at t = 0, v = v0 ]

or or

(3 t − t2 + v0 ) dt 3 t2 t 3 s= − + v0 t 2 3

∫ ds = ∫

26 | Mechanics-1 (a) Displacement at = Displacement at t = 0s t = 5s 3.52 53 0= − + v0 5 2 3 s ⇒ v0 = 5/6 = 0833 . (b) v = 3 t − t2 + v0 Velocity at t ( = 500 . s) = 3.5 − 52 + 0833 . = 15 − 25 + 833 = − 9.167 m/s 25. v = 3 t2 − 6t 2

∫ ds = ∫3(3 t

− 6t) dt s = t − 3 t2 + c s = t 3 − 3 t2 = t2 ( t − 3) (3.5)2 (3.5 − 3) (a) Average velocity = 3.5 = 3.5 × 05 . = 175 . m/s (b) Distance covered

∴ or or

Now, at t = 2 s, v = 6 m/s 62 ∴ = ( 4 × 2) + C 2 Thus, C = 10. v2 i.e., …(ii) = 4 t + 10 2 ∴ at t = 3 s, v2 = ( 4 × 3) + 10 2 v2 = 44 v = 44 = 2 11 m/s Substituting above found value of v in Eq. (i), 2 11 a = 4 2 i.e., a= 11 = 0.603 m/s 2 27. According to question, the velocity of the

particle varies as shown in figure. v 20 m/s

v

O

1 2

3 3.5

t

–3

=

2

2

3.5

∫ 0 (6t − 3 t ) dt + ∫ 2

(3 t2 − 6r) dt

= [3 t2 − t 3 ]20 + [ t 3 − 3 t2 ] 23.5 = 3(2)2 − (2) 3 + (3.5) 3 − 3(3.5)2 − (2) 3 + 3(2)2 = 12 − 8 + 42875 . − 3675 . − 8 + 12 = 14.125 m 14.125 (c) Average speed = 3.5 = 4.036 m/s 4 26. v= a i.e., …(i) va = 4 or ∫ v dv2 = ∫ 4 dt v or = 4t + C 2

30 m/s s

v s + =1 20 30 or 3v + 2s = 60 or …(i) 3v = 60 − 2s Differentiating above equation w.r.t. time t dv ds 3⋅ = − 2⋅ dt dt or 3a = − 2v 60 − 2 s or 3 a = − 2 ⋅   [using Eq. (i)]  3  2 or a = − (60 − 2 s) 9 2 [at s = 15 m] = − (60 − 2 × 15) 9 20 m/s2 =− 3

∴