Motion In Straight Line PDF

# MOTION IN STRAIGHT LINE - Distance → actual path - Displacement → shortest path - $\Delta n = n_f - n_i$ - Vector Quantity → It's direction - Distance ≥ |Displacement| - Distance = |Displacement| in case of straight line motion without turns. - Distance > 0 - Displacement > 0, = 0, < 0 **Que:**...

# MOTION IN STRAIGHT LINE - Distance → actual path - Displacement → shortest path - $\Delta n = n_f - n_i$ - Vector Quantity → It's direction - Distance ≥ |Displacement| - Distance = |Displacement| in case of straight line motion without turns. - Distance > 0 - Displacement > 0, = 0, < 0 **Que:** - Position of particle varies with time $n = t^2+t+1$ - Find displacement during 1st 2 seconds. - Find displacement during 3rd second. | Time | Position | |------|----------| | t=0 |$n_i = 1m$ | | t=2 | $n_f = 7m$ | | t=3 | $n_f = 13m$| - $n_f - n_i = 7-1 = 6m$ - displacement during 1st 2 seconds - $n_f - n_i = 13-7 = 6m$ - displacement during 3rd second ### Velocity - Displacement per unit time or rate of change of position. - It is a vector quantity - Its direction is parallel to direction of displacement - SI unit m/s ### Speed - Magnitude of velocity ### Uniform Velocity - Velocity = constant - Average velocity = instantaneous velocity ### Non-uniform velocity - Instantaneous velocity = V= $dy/dt$ - Average velocity = Total displacement/ total time ### Average Speed - Average speed = Total distance / Total time ### Instantaneous speed - Magnitude of instantaneous velocity **Que:** A particle is travelling straight line such that for half distance speed is $V_1$ and for remaining half distance is $V_2$, find average speed during entire motion. **Sol:** - $V_{avg} = (V_1 * x/2 + V_2 * x/2)/(x/2 + x/2)$ - $= V_1 + V_2/2$ - $V_{avg} = 2V_1V_2/(V_1 + V_2)$ **Que:** If a particle is moving in straight line such that for half time speed is $V_1$ and for remaining half time speed is $V_2$, find average speed for entire motion. **Sol:** - $V_{avg} = (V_1 * x/2 + V_2 * x/2)/ (x/2 + x/2)$ - $V_{avg} = V_1 + V_2/2$ **Que:** A particle is moving in a straight line such that for half distance of its motion speed is $V_1$ and for remaining half distance speed is $V_2$ for half time and $V_3$ for remaining half time, find average speed for entire motion. **Sol:** - $V_{avg} = 2 * V_1 * (V_2 + V_3)/(V_1 + V_2 + V_3)$ **Que:** Distance between A and B is 2.5 km. If a boy travels from A to B with speed 5 km/h and returns from B to A with speed 7.5 km/h, then find average speed and velocity during (t = 0 to t = 50 min) **Sol:** - Time taken from A to B = 5/5 = 50 min - Time taken from B to A = 2.5/7.5 = 20 min - At t = 50 min, the boy will be at A - Distance = 5 km - Average speed = (5/50 )* 60 = 6 km/h - Average velocity = 0 **Que:** Position of particle varies with time $n = (t^2 + t + 1)m$. Find velocity (i) During t = 0 to t = 2 sec - (ii) During t = 3 sec. | Time | Position | |------|----------| | t = 0 | $n_i = 1m$ | | t = 2 | $n_f = 7m$ | | t = 3 | $n_f = 13m$ | - Average velocity = $n_f - n_i / t = 7-1 / 2 = 6/2 = 3 m/s$ - $V_n = dn/dt = 2t +1$ - $V_n = 2 * 3 +1 = 7 m/s$ ### Average and Instantaneous Velocity from n-t graph - For uniform motion: - V = 1 - n = vt - y = mx - Slope = v = tan θ - Slope of n-t curve = velocity **Que:** Find $V_a/ V_b$ - $V_a = tan30 = 1/√3$ - $V_b = tan60 = √3$ - $V_a/V_b = (1/√3)/√3 = 1/3$ - $V_a : V_b = 1:3$ - For non-uniform motion: - Average velocity between A and B = slope of line AB ### Instantaneous Velocity - $tanθ = (n_2 - n_1) / (t_2 - t_1) = V_{avg}$ **Que:** - Position of the particle varies with time according to the relation $n = e^{-t} * t^2$. - Find time when velocity is zero. - $n = e^{-t} * t^2$ - $V = dn/dt = d(e^{-t} * t^2)/dt$ - $V = t^2(-e^{-t}) + e^{-t} * 2t$ - $V = 2te^{-t} - e^{-t} * t^2$ - $V = te^{-t} * (2-t)$ - For V = 0 - t = 0 or e^-t = 0 - Or t = 00 - 2 - t = 0, t = 2 sec - Thus t = 2 sec is the answer. **Que:** - At what point velocity of particle is equal to zero - (At B) - Find the points between velocity is equal to velocity of A and C at some line which is passing through A and C because slope will be zero at B - Velocity b/w point B and C will be zero. **Que: 7** n = $Ae^{-αt} + Be^{βt}$ - Find velocity of the particle and its nature, if time increases. - $V = dn/dt = d(Ae^{-αt} + Be^{βt})/dt$ - $V = -Ae^{-αt} + BBe^{βt}$ - If t is increasing, then β is increasing. - $V = -Ae^{-αt} + BBe^{βt}$ **Que:** Position of particle varies as $ n = t^2 - 4t + 2 $. Find displacement and distance during $t = 0$ to $t = 3 sec$ | Time | Position | |------|----------| | t = 0 | $n_i = 2$ | | t = 3 | $n_f = -1$ | - $n_f - n_i = -1 -2 = -3$ - displacement = -3 - Distance = -1 - 2 - Distance = -3 ### Acceleration - Rate of change of velocity - $a = \Delta v / \Delta t$ - It is a vector quantity. - It's direction is same as change in velocity. ### Uniform acceleration - $a = constant$ - Average acceleration = instantaneous acceleration ### Non-uniform acceleration - Instantaneous acceleration = $a = dv/dt$ ### Instantaneous acceleartion - When $V = f(x)$ - $a = dv/dx * dx/dt = dv/dx * dy/dt$ - $a = vdv/dx$ **Que:** If $V = \beta * √n$ find the initial velocity and acceleration of the particle. - $V = \beta * √n$ - squaring both sides - $V^2 = \beta^2 n$ - comparing with $V^2 = U^2 + 2as$ - $u= 0$ - $2a = \beta^2$ - $a = \beta^2 /2$ - $u=0$ **Que:** Find retardation: - $a = dv/dt$ - $dv = -2√n * \beta * dn$ - $dx = v dv = \beta * √n * -2√n * \beta * dn$. - $a * dx = -2\beta^2 * n * dn$ - $a = -2\beta^2n^{-1}$ **Que:** $t = 2√n + \beta n^2$ find acceleration. - Differentiate w.r.t 't' - 1 = $a * dx / dt + \beta^2 * dn / dt$ - 1 = $(a + 2 \beta^2 x) * dn / dt$ - 1 = $(a + 2 \beta^2 x) * V$ - $V = 1 / (a + 2 \beta^2 x)$ - $dv / dx = -1 * (a + 2\beta^2x)^{-2} * 2\beta^2$ - $a * dv / dn = -(a + 2\beta^2x)^{-3} *2\beta^2$ - $a [ (a + 2\beta^2x)^{-1} * 2\beta^2 ]$ - $a = -2\beta^2 * V^3$ ### Equations of motions: - $a=$ constant - $v = u + at$ - $s = ut + 1/2 * at^2$ - $v^2 = u^2 + 2as$ ### Increasing Speed - $v = u + at$ - $s = ut + 1/2 * at^2$ - $v^2 = u^2 + 2as$ ### Decreasing Speed - $v = u - at$ - $s = ut - 1/2* at^2$ - $v^2 = u^2 - 2as$ ### Displacement during tth second: - $S_{tth} = S_t - S_(t-1)$ - $S_{tth} = U + 1/2 * a * (2t - 1)$ ### Equation of motion under gravity - Upward Motion: - $v = u - gt$ - $h = ut-1/2* gt^2$ - $v^2 = u^2 - 2gh$ - $H_{th} = U - 1/2 * g * (2t -1)$ - At highest point: - Time to reach max height - $t= u/g$ - Maximum height - $H = u^2/2g$ - Downward Motion: - $-v = -u - gt$ - $-h = -ut -1/2 * gt^2$ - $v^2 = u^2 + 2gh$ **Que:** A particle is projected vertically upward with initial speed u under gravity. Find its speed at half of max height. - $h_{max} = v^2/2g$ - $v'^2 = v^2 - 2gh_{max}$ - $v'^2 = v^2 - v^2 = 0$ - $v' = v/√2$ - (a) $u=0, 1, 2, 3, 5$ - then $h_1 : h_2 : h_3 = 1:3:5$ - (b) $u ≠ 0$ - the $h_1 : h_2 : h_3 = 1: 3: 5$ - (c) $u ≠ 0$ , constant retardation - then $h_1 : h_2 : h_3 = 5: 3: 1$ - (d) $ u = 0$ - then $h_1 : h_2 : h_3 = 5: 3: 1$ **(e)** Water droplets are ejecting regularly: - 4th drop - 3rd drop - 2nd drop - 1st drop - n = h/x - Position of 2nd drop from the ground = 3h = 3* 5 = 15 - Position of 2nd drop from the ground = 3.75 m **Que:** Water droplets are ejecting from a tap at a height of 5m. If the 3rd drop is just ejecting when 1st drop hit the ground. Calculate the position of 2nd drop - x + 3h = x - s = wh - $h = 5m$ - Position of 2nd drop is = 3h = 3 * 5 = 15 = 3.75m - (a) $u = 0$ - then $t_1 : t_2 : t_3 = (√2 - 1) : (√3 - √2): (√5 - √3)$ - (b) $u ≠ 0 $ - then $t_1 : t_2 : t_3 = (√2 - 1) : (√3 - √2) : (2 - √3) : (√5 - 2)$ - (c) $u ≠ 0$ , constant retardation - then $t_1 : t_2 = (√2 - 1) : 1$ - (d) $ u = 0 $ - then t1: t2: t3 = 5: 3: 1 **Que:** Particle is moving with some initial speed u, and constant retardation a, if its total time of the motion is T what will the time for 1st half distance? - $s = ut + 1/2 at^2$, u = 0 - $T =t_1 + t_2$ - $t_1 + t_2 = (√2 -1)* t + t$ - $T = √2 * t$ - $t = T/√2$ - Time for 1st half $= (√2 - 1)* T/√2$ - Time for 2nd half $= T/√2$ **Que:** 3. $u=0, S_1, V_{max}, S_2, V =0 $ - A, B = retardation - (1) Total time for motion = T - Then maximum speed and total distance covered for AB - $V = u + at$ - $V_{max}= 0 + at_1$ - $t_1 = V_{max}/ a$ - For BC - $V = u + at$ - $0 = V_{max} - \beta *t_2$ - $t_2 = V_{max}/\beta$ - $t = t_1 + t_2$ - $t = V_{max} (1/a +1/ \beta )$ - $V_{max} = a \beta t / (a + \beta)$ - $S_1 = (0 + V_{max}) * t_1$ - $S_2 = (V_{max} + 0) * t_2$ - $S = a \beta t^2 / (a + \beta)$ ### Reaction time and stopping distance: - Reaction time: Time between observation of an obstacle and break applied. - During reaction time the particle/object will move with same velocity: - $S_2 =$ stopping distance - $V^2 = U^2 - 2as$ - $S_2 = U^2/2a$ - (ii) when air resistance = 0, $t_1 = t_2$ - (iii) when air resistance ≠ 0, $t_1 < t_2$ - $t_1$ - time of ascending - $t_2$ - time of descending **Que:** A ball is projected vertically up under gravity. If the ball travels equal distances during 5th second and 8th second of its motion, then find the maximum height travelled by the ball. - t = 7 - t = 8 - t = 9 - t = 10 - Total time = 10 sec - Time of ascent = 5 sec. - $u = sg$ - $h_{max} = u^2/2g$ - $u = 5 * 10 = 50 m/s$ - $h_{max} = 50^2/2 * 10$ - $h_{max} = 125 m$ **Que:** Distance of height covered during last 't' second of upward motion - $h = ut + 1/2 * gt^2$ - $h = 1/2 * gt^2$ - Distance in last 't' second in upward motion = $1/2* gt^2$ **Que:** A particle is projected vertically up with speed 45 m/s under gravity. What is the distance travelled by particle during 5th second of its motion *g = 10 m/s^2* - $V = u - gt$ - $0 = 45 - 10 * t$ - $t = 4.5$ - $h = -1/2 * g * (2t - 1)$ - $h = 45 - 10 * (2 * 5 -1)$ - $h = 0$ - Distance = 1.25 m - Total height covered = 1.25m - Distance = 1.25 + 1.25 = 2.5 m **Que:** A particle is projected with speed 37 m/s vertically up under gravity. Find distance and displacement covered by particle during 4th second of its motion. - V = u-gt - $0 = 37 - 10 * t$ - $t = 3.7$ - $h = 1/2 * gt^2$ - $h = 1/2 * 10 * (0.3)^2$ - $h = 0.145$ - Distance = $h_1 - h_2$ - Distance = 2.45 + 0.45 = 2.9m - Displacement = $h_1 – h_2$ - Displacement = 2.45 - 0.45 = 2m **Que:** A particle is dropped from a height ‘H’. If it travels 5hm in the last 2 sec of its motion, then find the value of ′H′ (*g = 10*). - $h = 1/2 * gt^2 $ - $4 = 1/2 * 10 *(t-2)^2$ - $4 = 5 *(t-2)^2$ - $√4/5 = (t-2)$ - $2/√5 = (t-2)$ - $t = 2 + 2/√5$ - $t = (2√5 + 2)/√5$ - $t = 3.8 + 2 = 5.8$ - $h = 1/2 *gt^2$ - $h =1/2 * 10 * (5.8)^2$ - $h = 180m$ **Que:** A ball is projected vertically up from a tower of height 100m with speed 50 m/s under gravity. Calculate the time taken by the ball to reach the ground. - $1/2 * gt^2 - ut - h = 0$ - $1/2 * 10 * t^2 - 20 * t - 100 = 0$ - $5t^2 - 20t - 100 = 0$ - $5(t^2 - 4t -100) = 0$ - $t = 4±√16+400/2$ - $t = 4±√416/2$ - $t = 4 ± 20.4/2$ - $t = 12.2$ - $t = -8.2$ - $t = 6.9 sec$ **Que:** A balloon is rising up with a constant velocity 4.9 m/s. When it is at the height of 9.8m. Stone is dropped from it. Calculate time taken by stone to reach at ground. *(g = 9.8 m/s^2)* - $1/2 * gt^2 - ut - h = 0$ - $1/2 * 9.8 * t^2 - 4.9 * t - 9.8 = 0$ - $4.9t^2 - 4.9t - 9.8 = 0$ - $t^2 - t - 2 = 0$ - $t= 1 ± √1 + 8/2$ - $t = 1 ± 3/ 2$ - $t = 2 sec$ **Que:** If a particle is seen twice at the same height: - $h = ut + 1/2 * gt^2$ - $h = ut + 1/2 * gt^2$ - $t_1 + t_2 = -b/a$ - $t_1 * t_2 = c/a$ - $(t_2 - t_1)^2 = (t_1 + t_2)^2 - 4t_1t_2$ - $t_2 - t_1 = √(t_1 + t_2)^2 - 4t_1t_2$ - $t_2 - t_1 = 2 * √u^2/g - 2gh/g$ ### Relative Velocity: - Car -> V - Boy -> - Velocity of car w.r.t boy = $V_{cb} = V$ - Velocity of boy w.r.t car = $V_{bc} = - V$ - (a) When two objects are moving in the same direction: - $V_{ab} = V_a - (V_b)$ - $V_{ab} = V_a - V_b$ - $V_{ba} = V_b - V_a$ - (b) When objects are moving in the opposite direction: - $V_{ab} = V_A - (-V_b)$ - $V_{ab} = (V_a + V_b)$ - $V_{ba} = (V_b - V_a)$ - Velocity B w.r.t A: - $V_{ba} = V_b - (-V_a)$ - $V_{ba} = V_b + V_a$ - $V_{ab} = - V_{ba}$ ### Relative Acceleration: - $a_{ab} = (a_a - a_b)$ - $a_{ba} = (a_b - a_a)$ - $a_{ab} = a_a - (-a_b) = (a_a + a_b)$ - $a_{ba} = a_{ba} - (-a_a)$ - $a_{ba} = (a_{ba} + a_a)$ ### Downstream: - $V_w = $ velocity of water w.r.t ground. - $V_s = $ velocity of swimmer w.r.t still water - $V_{sg}$ = $V_s + V_w$ ### Upstream: - $V_{sg} = V_s - V_w$ **Que:** Ram takes $T_1$ time to cover up a certain distance on an escalator. If the escalator is at rest and Ram takes $T_2$ time to cover the same distance on an escalator when it is moving. What time will it take for him if both are at rest? - $V_e = 1/t_1$ - $V_r = 1/t_1$ - $V_{re} = V_e + V_r$ - $1/t = 1/t_1 +1/t2$ - $t = (t_1 * t_2) / ( t_1 + t_2 ) $ **Que:** A lift is moving downward with acceleration ‘a’. A ball is dropped in it. Find acc of ball w.r.t ground and acc of ball w.r.t lift. - Acc of ball w.r.t. ground = g = 1 - Acc of ball w.r.t. lift = g- a **Que:** When car A is ‘d’ distance behind the car B, then car A start to drive with ‘a’ find the condition such that no collision will occur. **Sol:** - Initial velocity of ‘A’ w.r.t B: $V_{ab} = V_A - V_b$ - Final velocity of ‘A’ w.r.t ‘B’ at the time of collision will be zero - retardation of ‘A’ with ‘B’ - $a_{ab} = - a - 0 = -a$ - $V_{ab}^2 = U_{ab}^2 - 2as$ - $0 = (V_A - V_B)^2 - 2as$ - $S = (V_A - V_B)^2/2a$ - For no collision - $d ≥ S$ - $d ≥ (V_A - V_B)^2 / 2a$ **Que:** A bus starts with a constant acceleration 2 m/s^2. Calculate the time when (1) a boy catches to bus (t_1) and (2) a boy catches to boy (t_2) **Sol:** $a = 2 m/s^2$ - Let after time ‘t’ boy catches to bus. - $S_1 = 10 + S$ - $10t = 10 + 1/2 *2 *t^2$ - $10t = 10 + t^2$ - $t^2 -10t + 10 =0$ - $t = 10 ± √10^2 - 40 /2$ - $t = 10 ± √60/2$ - $t_1 = 10 - √60/2 = 1.1 sec$ - $t_2 = 10 + √60/2 = 8.9 sec$ **Que:** In the previous problem, the boys were winning with speed 10 m/s instead of 10 m/s. What the value of V such that that boy catches the bus. - $S = 10 + S$ - $Vt= 10 + 1/2 * 2* t^2$ - $Vt = 10 + t^2$ - $t^2 - Vt + 10 = 0$ - $V^2 - 4* 1 * 10 ≥ 0$ - $V^2 ≥ 40$ - $V ≥ √40$ - $V ≥ 6.32$ - to catch the bus. ### Graphical Representation: - Time to meet: - Relative velocity: $V_1 + V_2$ - Relative acceleration = g -g = 0 - $t = h/ (V_1 + V_2)$ - Graph: y- axis - $dy/dx = 0$: - $dy/dx = 0$ - slope of y-x curve = $dy/dx$ - for maxima and minima. - If $dy/dx = 0$ and, - $d^2y/dx^2 > 0$ - minima. - $d^2y/dx^2 < 0$ – maxima. **Que:** If $y = 3x^2 - 6x + 10$, find the value of ‘x’ for the maximum and minimum value of ‘y’. - $y = 3x^2 - 6x + 10$ - $dy/dx = 6x - 6 = 0$ - $x = 1$ - $d^2y/dx^2 = 6 > 0$ - minima. - $x = 1; y = minima$ - $y_{min} = 3 - 6 + 10 = 7$ **Que:** $y = √3 sin θ + cos θ$, find the minimum value of ‘y’. - $y = √3 sinθ + cosθ$ - $a = √3$ - $b = 1$ - $√a^2 + b^2 = √3^2 + 1^2 = 2$ - $y = 1/2 * (sin θ cos 30^o + cos30^o sin θ)$ - $y = 1/2 * sin ( θ + 30^o)$ - min - $sin ( θ + 30^o )= max = 1$ - $y_{min} = 1/2$ - $y = 1/2 * (√3 sin θ + cos θ)$ - Let $P =√3sin θ + cos θ$ - If y = min - P = max - $dP/dθ = √3cosθ - sin θ = 0$ - $tan θ = √3$ - $θ = 60^o$ - $y_{min} = √3 sin 60^o + cos 60^o$ - $y_{min} = √3 * √3/2 + 1/2$ - $y_{min}= 3/2 + 1/2 = 1$ - Straight line: - $y = mx + c$ - $m = tan θ$ - Parabola: - (i) $y^2 = kn$ - (ii) $x^2 = ky$ - (iii) $y = ax + bx^2$ where - a ≠ 0, b ≠ 0 - $x = 0$ - $n = a/b$ - (iv) $ y = -ax + bx^2$ where - $x = 0$ - $ n= a/b$ - $b < 0$ - (v) $y = +ax - bx^2 $ - Ellipse: - a = semi major axis - b = semi minor axis - $e = √(1-(b^2/a^2))$ - $e<1$ for ellipse. - Area of ellipse = A = $ πab$ - Uniform motion: - Velocity = constant - $V = 1$ - $m = 1/t$ - $V = tan θ = y/t$ - (i) For (-ve ) velocity - $tan θ < 0$ - $θ > 90°$ - (ii) For Speed - $ θ < 90°$ - $ V > 0$ - (iii) For zero velocity - $V = 0$ - $ θ = 0$ ### Velocity - time for uniform motion: - (i) $V > 0$ - (ii) $V < 0$ - (iii) $V = 0$ - Area of velocity - time = Displacement. - Area of speed - time = Distance. ### Any graph which is indicating negative speed, is not practically possible. - A graph indicating uniform acceleration motion. - $V = ut + at$ - $v= at + u$ - $y = mx + c$ where - $a = v - u /t = tan θ $ - Slope of v-t curve = acceleration - (i) $a < 0$ and $v = 0$ - (ii) $ a > 0$ - (iii) $a = 0$ - (a) a > 0 - (b) a < 0 - Area of a-t curve change in velocity. ### a > 0 - Increasing speed - Slope of n-t curve = velocity - Slope of v-t curve = acceleration - Area of velocity - time = Displacement - Area of a-t curve = change in velocity. ### a < 0 - Decreasing speed - Slope of n-t curve = velocity - Slope of v-t curve = acceleration - Area of velocity - time = Displacement - Area of a-t curve = change in velocity. **Que:** - A particle is projected with velocity v vertically up under gravity. Calculate Draw v-t (velocity - time), a-t, n-t (position - time) - $v = v - gt$ - $v = -gt + v (speed-time graph)$ - $a = -g$ - $a = -g$ - Distance-time graph: - a < 0 **Que:** - $n = 20 m$ - $t = 0, 6, 8$ - (i) find average speed t = 0 -8 sec - (ii) find velocity at t = 1, t = 5, t = 7 sec - (iii) find acceleration t = 0 to t = 8 sec - Distance = 20 + 0 + 20 = 40m - Average velocity = 40 m/ 8 = 5 m/s - (i) V = 20 = 10 m/s, at t = 1 - (ii) V = 10 m/s at t = 5 - (iii) V = 10 m/s - at t = 7 - Average acceleration = (final velocity - initial velocity)/time = ( - 10 - 10 ) / 8 = -20 / 8 = - 2.5 m/s^2 **Que:** - $ n = 0 m$ - If particle starts with velocity V = - 10m/s, then find velocity at t = 6 sec. - Area = $l * b + 1/2 * b * h$ - Area = $20 * 4 + 1/2 * 2 * 1 * 20$ - Area = $ 100$ - $V - (-10) = 100$ - $V = 90 m/s$ **Que:** - $Vo$ - Find a-n curve. - $a = dv/dt = (-V_o * x / V_o * x) /