Tiêu đề 3.MOTION IN A STRAIGHT LINE - Explanations.pdf ✅

1 (c) For shortest possible path man should swim with an angle (90 + θ) with downstream From the fig, sin θ = vr vm = 5 10 = 1 2 ⇒ ∴ θ = 30° So angle with downstream = 90° + 30° = 120° 2 (d) From first equation of motion, we have v = u + at Given, u = 0, a1 = 2 ms−2 t = 10 s, ∴ v1 = 2 × 10 = 20 ms−1 In the next 30 s, the constant velocity becomes v2 = v1 + a2t2 Given, v1 = 20 ms−1 , a2 = 2 ms−2 ,t2 = 30 s ∴ v2 = 20 + 2 × 30 = 80 ms−1 . When it decelerates, then v3 2 = u 2 − 2a3s Here, v3 = 0 (train stops), v2 = 80 ms−1 , a3 = 4 ms−2 0 = (80) 2 − 2 × 4 × s Or s = 80 × 80 8 = 800 m. 3 (b) From equation of motion, we have v = u + gt Taking downward direction negative u = 10 + 5 = 15ms−1 , g = 10 ms−2 ,t − 2 s ∴ v = 15 − 2 × 10 = −2 ms−1 4 (a) Using V = u + at V = gt ...(i) Comparing with y = mx + c Equation (i) represents a straight line passing through origin inclined x-axis (slope -g) 5 (a) x = 1 2 gt 2 , 100 − x = 25x − 1 2 gt 2 ; Adding 25t = 100 or t = 4 s 6 (a) Distance covered in 5 th second S5 th = u + a 2 (2n − 1) = 0 + a 2 (2 × 5 − 1) = 9a 2 and distance covered in 5 second, S5 = ut + 1 2 at 2 = 0 + 1 2 × a × 25 = 25a 2 ∴ S5 th S5 = 9 25 7 (b) Here v = 144 km/h = 40m/s v = u + at ⇒ 40 = 0 + 20 × a ⇒ a = 2 m/s 2 ∴ s = 1 2 at 2 = 1 2 × 2 × (20) 2= 400 m 8 (b) At maximum height velocity v = 0 We know that v = u + at, hence 0 = u − gT ⇒ u = gT When v = u 2 , then u 2 = u − gt ⇒ gt = u 2 ⇒ gt = gT 2 ⇒ t = T 2 Hence at t = T 2 , it acquires velocity u 2 9 (b) S2 = 1 2 gt2 2 = 10 2 × (3) 2 = 45 m S1 = 1 2 gt1 2 = 10 2 × (5) 2 = 125 m ∴ S1 − S2 = 125 − 45 = 80 m 10 (d) Using, v = u + at or v − u = at, we find that if |a⃗ | is, it t is the time for acceleration, then t 2 is the time for retardation Now, t + t 2 = 3 or 3t 2 = 3 or t = 2s S = 1 2 × 2 × 2 × 2 + 1 2 × 4 × 1 × 1 = (4 + 2)m=6m 11 (a) Hmax ∝ u 2 ∴ u ∝ √Hmax i. e. to triple the maximum height, ball should be thrown with velocity √3 u 12 (d) vm W E  vR vR
Net displacement = 0 and total distance = OP + PQ + QO = 1 + 2π × 1 4 + 1 = 14.28 4 km Average speed = 14.28 4×10/60 = 6 × 14.28 4 = 21.42 km/h 13 (d) Relative velocity = 10 + 5 = 15 m/sec ∴ t = 150 15 = 10 sec 14 (a) vrel = 45 + 36 = 81 kmh−1 = 81 × 5 18 ms−1 srel = vrel × t = 81 × 5 18 × (5 × 60) = 81 × 5 × 5 × 60 18 − 6750 m = 6.75 km 16 (b) Let t1,t2 and t3 be the time taken by the particles to cover the distance 2x, 4x and 6x respectively. Let v be the velocity of the particle at B ie, maximum velocity. The particle moves with uniform acceleration from A to B. For motion for A to B. Average velocity = 0+v 2 = v 2 Time taken, t1 = 2x v/2 = 4x v Particle moves with uniform retardation from C to D. Time taken , t3 = 6x (0+v)/2 = 12x v Total time = t1 + t2 + t3 = 4x v + 4x v + 12x v = 20x v vav = 2x + 4x + 6x 20x/v = 12v 20 or v v = 12 20 = 3 5 . 17 (b) Average velocity = Total distance covered Time of flight = 2Hmax 2u/g ⇒ vav = 2u 2 /2g 2u/g ⇒ vav = u/2 Velocity of projection = v [Given] ∴ vav = u/2 18 (b) Total time of motion is 2 min 20 sec = 140 sec As time period of circular motion is 40 sec so in 140 sec. Athlete will complete 3.5 revolution i. e., He will be at diametrically opposite point i. e.,Displacement = 2R 19 (c) From acceleration time graph, acceleration is constant for first part of motion so, for this part velocity of body increases uniformly with time and as a = 0 then the velocity becomes constant. Then again increased because of acceleration 20 (b) Let us solve the problem in terms initial velocity, relative acceleration and relative displacement of the coin with respect to floor of the lift. u = 10 − 10 = ms −1 , a = 9.8ms −2 , S = 4.9m,t =? 4.9 =0× t + 1 2 × 9.8 × t 2 or 4.9t 2 = 4.9 or t = 1s 15 = 30 − 10t or 10t = 15 or t = 1.5s 21 (c) dx dt = 4t 3 − 2t or dx = 4t 3dt − 2t dt Integrating, x = 4t 4 4 − 2t 2 2 = t 3 − t 2 When x = 2,t 4 − t 2 − 2 = 0, t 2 = −(−1) ± √1 + 8 2 or t 2 = 1±3 2 = 2 (ignoring – ve sign) Again, d 2x dt 2 = 12t 2 − 2 When t 2 = 2, acceleration=12× 2 − 2 = 22ms −2 22 (b) The distance travelled in t sec in upward motion is s = u − 1 2 g(2t − 1) ∴ AB = u − 1 2 g(2 × 5 − 1)
AB = u − 1 2 9g Distance travelled in 1 s in the downward direction is BA = 0 + 1 2 g(1) 2 It is given that these distance are equal. Therefore, u − 9g 2 = 1 2 g ⇒ u = 5 × 9.8 = 49 ms−1 23 (b) Let height of minaret is H and body take time T to fall from top to bottom H = 1 2 gT 2 ...(i) In last 2 sec body travels distance of 40 m so in (T − 2) sec distance travelled = (H − 40)m (H − 40) = 1 2 g(T − 2) 2 ...(ii) By solving (i) and (ii), T = 3 sec and H = 45m 24 (c) Let the man starts crossing the road at an angle θ with the roadside. For safe crossing, the condition is that the man must cross the road by the time truck describes the distance(4+2 cot θ), So, 4+2cotθ 8 = 2lsinθ v or v = 8 2sinθ=2cosθ For minimum v = dv dθ = 0 or −8(2cosθ−sinθ) (2sinθ+cosθ)2 = 0 or 2cosθ − sinθ = 0 or tanθ = 2, so, sinθ = 2 √5 , cosθ= 1 √5 ∴ vmin = 6 2 ( 2 √5 ) + 1 √5 = 8 √5 = 3.57ms −1 25 (a) S1 = 1 2 ft 2 , S2 = −v0t − 1 2 gt 2 , Clearly, (S1 − S2) ∝ t 26 (a) When the stone is released from the balloon. Its height h = 1 2 at 2 = 1 2 × 1.25 × (8) 2 = 40 m and velocity v = at = 1.25 × 8 = 10 m/s Time taken by the stone to reach the ground t = v g [1 + √1 + 2gh v 2 ] = 10 10 [1 + √1 + 2 × 10 × 40 (10) 2 ] = 4sec 27 (d) In the positive region the velocity decreases linearly (during rise) and in the negative region velocity increases linearly (during fall) and the direction is opposite to each other during rise and fall, hence fall is shown in the negative region 28 (c) Given : Initial velocity of a body u = 0 ... (i) Let s be the distance covered by a body in time t ∴ s = ut + 1 2 at 2 or s = 1 2 at 2 [Using (i)] ⇒ s ∝ t 2 29 (b) u = 0, v = 180 km h −1 = 50 ms −1 Time taken t = 10s a = v − u t = 50 10 = 5 ms −2 ∴ Distance covered S = ut + 1 2 at 2 = 0 + 1 2 × 5 × (10) 2 = 500 2 = 250 m 30 (b) Distance covered =Area enclosed by v − t graph = Area of triangle = 1 2 × 4 × 8 = 16 m 31 (c) x = at + bt 2 − ct 3 , a = d 2x dt 2 = 2b − 6ct 32 (b) Power, P = W t P = Fs t , P = mas t (∵ F = ma) P = mv s t 2 , (∵ a = v t ) P = ms∙s t 3 (∵ v = s t ) Pt 2 = ms 3 ∴ s ∝ t 3/2 33 (d) If t1 and t2 are the time, when body is at the same height then, h = 1 2 gt1t2 = 1 2 × g × 2 × 10 = 10 g 34 (d) In ‘s-t’ graph (positive -time) The straight line parallel with time axis represent state of rest 35 (a) tan(90° − θ)= 20 15 H H (H-40) m 40m 2 sec (T-2)sec T H
∴ cotθ = 20 15 = 4 3 ⇒ θ = 37° ∴ θ = 37° + 23° = 60° 36 (b) t = √ 2h g ⇒ t1 t2 = √ h1 h2 37 (a) Sn = u + g 2 (2n − 1); when u = 0, S1: S2: S3 = 1: 3: 5 38 (c) v 2 = u 2 + 2gh ⇒ v = √u 2 + 2gh So for both the cases velocity will be equal 39 (b) Time of ascent = Time of descent=5sec 40 (a) s ∝ t 2 [Given] ∴ s = Kt 2 Acceleration a = d 2 s dt 2 = 2k [constant] It means the particle travels with uniform acceleration 41 (c) When packet is released from the balloon, it acquires the velocity of balloon of value 12 m/s. Hence velocity of packet after 2 sec, will be v = u + gt = 12 − 9.8 × 2 = −76 m/s 42 (b) x = 9t 2 − t 3 ; v = dx dt = 18t − 3t 2 , For maximum speed dv dt = d dt [18t − 3t 2 ] = 0 ⇒ 18 − 6t = 0 ∴ t = 3 sec i.e., Particle achieve maximum speed at t = 3 sec. At this instant position of this particle, x = 9t 2 − t 3 = 9(3) 2 − (3) 3 = 81 − 27 = 54 m 43 (d) Up to time t1 slope of the graph is constant and after t1slope is zero i. e. the body travel with constant speed up to time t1 and then stops 44 (b) For vertically upward motion, h1 = v0t − 1 2 gt 2 and for vertically downward motion, h2 = v0t + 1 2 g t 2 ∴ Total distance covered in t sec h = h1 + h2 = 2v0t 45 (c) Acceleration = a = dv dt = 0.1 × 2t = 0.2t Which is time dependent i. e. non-uniform acceleration 46 (c) Slope is negative at the point E. 47 (a) t = √ 2h g ⇒ t1 t2 = √ h1 h2 = √ 1 2 = 1 √2 48 (d) The straight vertical line in the graph represents change in the direction of velocity. 49 (b) Let the particle moves toward right with velocity 6 m/s. Due to retardation after time t1 its velocity becomes zero From v = u − at ⇒ 0 = 6 − 2t1 ⇒ t1 = 3sec But retardation work on it for 4 sec.It means after reaching point A direction of motion get reversed and acceleration works on the particle for next one second. SOA = ut1 − 1 2 at1 2=6 × 3 − 1 2 (2)(3) 2 = 18 − 9 = 9m SAB = 1 2 × 2 × (1) 2 = 1m ∴ SBC = soa − saB = 9 − 1 = 8m Now velocity of the particle at pint B in return journey v = 0 + 2 × 1 = 2m/s In return journey from B to C, particle moves with constant velocity 2m/s to cover the distance 8m. Time taken = Distance Velocity = 8 2 = 4 sec Total time taken by particle to return at point O is ⇒ T = tOA + tAB + tBC = 3 + 1 + 4 = 8 sec 50 (c) For first part, u = 0,t = T and acceleration = a ∴ v = 0 + aT = aT and S1 = 0 + 1 2 aT 2 = 1 2 aT 2 For Second part, u = aT, retardation = a1, v = 0 and time taken = T1 (let) ∴ 0 = u − a1T1 ⇒ aT = a1T1 And from v 2 = u 2 − 2aS2 ⇒ S2 = u 2 2a1 = 1 2 a 2T 2 a1 S2 = 1 2 aT × T1 (As a1 = aT T1 ) ∴ vav = S1 + S2 T + T1 = 1 2 aT 2 + 1 2 aT × T1 T + T1 = 1 2 aT(T + T1) T + T1 = 1 2 aT 51 (c) For no collision, the speed of car A should be reduced to vB before the cars meet, ie, final relative