Uniformly Accelerated Motion
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2 Uniformly Accelerated Motion Accelerated Motion A railway train starting from a station gradually picks up speed. After some time when it has acquired sufficient speed it runs with a constant speed. Finally, it is brought to rest at another station by the application of brakes. During this interval speed of the train was gradually decreasing. The motion of the train when it started from first station and when it was nearing the second is said to be accelerated motion. Acceleration Acceleration is defined as the change in velocity in one second or Acceleration is defined as the rate of change of velocity Acceleration is a vector quantity. Its units in C.G.S. system is cms - system ms -. and in M.K.S. Direction of acceleration is along the direction in which change in velocity is represented. The motion may be uniformly accelerated motion or it may be non-uniformly accelerated, depending how the velocity changes with time. Direction of the Acceleration Vector Acceleration is a vector quantity so it will always have a direction associated with it. The direction of the acceleration vector depends on two factors: whether the object is speeding up or slowing down whether the object is moving in the positive (+) or negative ( ) direction The general RULE OF THUMB is: If an object is slowing down, then its acceleration is in the opposite direction of its motion. This RULE OF THUMB can be applied to determine whether the sign of the acceleration of an object is positive or negative, right or left, up or down, etc. We are often interested in how fast the velocity is changing. This is the acceleration.
3 Acceleration is a change of velocity divided by a change of time so it has units of (meters/second)/second. We will write this as m/s/s or m/s (there are no "square seconds"). (i) Uniform acceleration The acceleration of a body is said to be uniform if its velocity changes by equal 'amounts in equal intervals. (ii) Non-uniform acceleration The acceleration of a body is said to be non-uniform if its velocity changes by unequal amounts in equal intervals of time. (iii) Instantaneous acceleration The acceleration of a body measured at any instant is called instantaneous acceleration. a = dv / dt (iv) Average acceleration : It is defined as ratio of total change in velocity to the total time. Terminology u Initial velocity v Final velocity a Acceleration t Time x 0 Initial position x Position after time t a av = total change in velocity / total time Equations of Motion by Calculus Method Velocity Time Relation Acceleration, a = dv dt or dv = adt Integrating the above, v u = at v = u + at
4 Displacement Time Relation Instantaneous velocity, or dx = vdt dx v = dt dx = (u + at) dt [From velocity time relation] Integrating the above relation, t x = u + a x0 0 1 (x x 0) = ut + at x t [ ] [ t] 1 at (x x 0) = ut + Velocity Displacement Relation t 0 adx = vdv Integrating the above expression, x [ ] x0 a x v = a[ x x ] v u v u 0 = v u = a (x x 0 ) Putting (x x 0 ) = S, we obtain [S is the displacement] v u = as Graphical Method: (i) v = u + at (ii) S = ut + ½ at. (iii) V -u = as graphically (i) Slope of v-t graph gives acceleration.
5 Bc u a= tanθ = = V Ac t u a = v v u = at t v = u + at (ii) 1 S = ut + at A = Area ABC + Area ACDOA 1 = x AC BC + ( AO)( OD) 1 = tv ( u) + ( u)( t) 1 S = ( t )( at ) + ut as v u = at 1 ( ) S = ut + at (iii) v u = as S = area OS Trap. OABDO 1 = ( OA + BD )( AC ). 1 = ( u+ v )( t ) 1 v u v u S = ( u+ v ) = t. a a as = v u = v u as Motion of an Object under Free Fall If we choose upward direction as positive, then gravitational acceleration g will be taken negative. a = g = 9.8 m s As object is released from rest, therefore, u = 0 The equation of motion becomes v = 0 gt = 9.8t m s 1
6 y = 0 gt = 4.9t m v = 0 gy = 19.6y m s Note Distance travelled in n th second of uniformly accelerated motion is given by the relation, Problems Based on Equations of Motion for Uniformly Accelerated Motion A particle starts with an initial velocity 5.0 ms 1 along the positive x-direction and it accelerates uniformly at the rate of ms. (i) Find the distance travelled by it in the first three seconds. (ii) How much time does it take to reach the velocity 9.0 ms 1? (iii) How much distance will it cover in reaching the velocity 9.0 ms 1? Solution Here, u = 5.0 ms 1 a = ms (i) We have to calculate S, when t = 3 s (ii) We have to calculate t, when v = 9 ms 1 v= u + at 9 = 5 + () t 9 5 = t t = 4 t = s (iii) Here, v = 9 m s, S =? v = u + as (9) = (5) + S 81 = 5 + 4S 4S = 81 5
7 4S = 56 S = 14 m Problems Based On Free Fall of a Body A ball is thrown vertically upwards with a velocity of 30 ms 1 from the top of a multistoreyed building. The height of the point from where the ball is thrown is 0.0 m from the ground. (i) How high will the ball rise? (ii) How long will it be before the ball hits the ground? Take g = 10 ms Solution (i) For vertical motion, u = 30 ms 1 a = 10 ms (Upward motion) v = 0 (Highest point) S =? t = t 1 (say) As v = u + as, 0 = (30) + ( 10) S S = 45 m As v = u + at, 0 = 30 + ( 10) t 1 t 1 = 3 s (ii) For vertical downward motion, u = 0 ms 1 a = 10 ms S = (45 + 0) m = 65 m t= t (say) 1 = +, As S ut at t 65 = = 13 s 5 t = 13 s = 3.6s Total time = t 1 + t = ( ) s = 6.6 s
8 Relative Velocity Relative Velocity The relative velocity of a body A with respect to another body B v AB is the time rate at which A changes its position with respect to B. Case 1: Both bodies move in the same direction If A and B are moving in the same direction, then the resultant relative velocity is Case : The bodies move in opposite directions If A and B are moving in the opposite directions, then the resultant relative velocity is v = v v AB A B Example 1 Two cars X and Y are moving with speeds of 60 km h 1 and 80 km h 1 respectively along parallel straight paths. Both the cars started from the same position. What is the position of car X with respect to Y after 15 minutes? Solution Speed of car X, v x = 60 km h 1 Speed of car Y, v y = 80 km h 1 Both the cars are moving in the same direction. Thus, relative velocity of car X with respect to car Y, v xy = v x v y = (60 80) km h 1 = 0 km h 1 Separation of car X with respect to car Y after 15 minutes
9 is = 5km [ Distance = Speed Time] Example Two trains, 110 m and 90 m in length, are running in opposite directions with velocities 75 km h 1 and 64 km h 1. In what time will they completely cross each other? Solution Here, v A = 75 km h 1 v B = 64 km h 1 [Trains are moving in opposite directions] Length of train A, l A = 110 m Length of train B, l B = 90 m Relative velocity of the two trains is v AB = v A v B = 75 ( 64) = ( ) km h 1 = 139 km h 1 = 139 km h 1 = 38.6 m s 1 Total distance to be travelled by each train for completely crossing the other train = ( ) = 00 m 00m = 5.s 38.6ms Time taken by each train to cross the other train 1
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