Kinematics Solved Example

KINEMATICS Page # 12 SOLVED EXAMPLE Ex.1 A car is moving along a straight line, say OP in Fig. 3.1. It moves from O to P in 18 s and returns from P to Q in 6.0 s. What are the average velocity and average speed of the car in going (a) from O to P ? and (b) from O to P and back to Q ? Ans. (a) Average velocity =   Displacement Time interval  360 m  20 m s 1 18 s Average speed  Path length 360 m  20 m s1 = Time int erval 18 s Thus, in this case the average speed is equal to the magnitude of the average velocity. (b) In this case, Average velocity = Displacement 240 m = Time interval (19  6.0) s = + 10 ms–1 Path length OP  PQ Average speed  = Time int erval t (360  120) = = 20 m s–1 24 s Thus, in this case the average speed is not equal to the magnitude of the average velocity. This happens because the motion here involves change in direction so that the path length is greater than the magnitude of displacement. This shows that speed is, in general, greater than the magnitude of the velocity. ?? If the car in Example 3.1 moves from O to P and comes back to O in the same time interval, average speed is 20 m/s but the average velocity is zero ! Ex.2 The position of an object moving along x-axis is given by x = a + bt2 where a = 8.5 m, b = 2.5 m s–2 and t is measured in seconds. What is its velocity at t = 0 s and t = 2.0 s. What is the average velocity between t = 2.0 s and t = 4.0 s ? Ans. In notation of differential calculus, the velocity is   dx (a  bt 2 )  2b t  5.0m s 1 dt At t = 0 s,  = 0 m s–1 and at t = 2.0 s.  = 10 m s–1, Average velocity = x(4.0)  x(2.0) 4.0  2.0 a  16b  a  4b = 6.0 × b = 6.0 × 2.5 = 15 m s–1 2 .0 From Fig. 3.7, we note that during the period t =10 s to 18 s the velocity is constant. Between period t =18 s to t = 20 s, it is uniformly decreasing and during the period t = 0 s to t = 10 s, it is increasing. Note that for uniform motion, velocity is the same as the average velocity at all instants. Ex.3 Obtain equations of motion for constant acceleration using method of calculus. Ans. By definition a  d dt d = a d t Integrating both sides   d  0  t a dt 0  a t 0 dt (a is constant)  – 0 = at,  = 0 + at Further, dx , dx =  dt dt = Integrating both sides  x x0 dx  t t 0 0   dt   ( x  x0  0t  0  at) dt 1 a t2 2 We can write a d d dx d   or,  d = a dx dt dx dt dx Integrating both sides,   0 d   x x0 adx , 2  20  a(x  x0 ) 2 2  20  2a(x  x0 ) The advantage of this method is that it can be used for motion with non-uniform acceleration also. Now, we shall use these equations to some important cases. KINEMATICS Page # 13 Ex.4 A ball is thrown vertically upwards with a Ex.6 Find the magnitude and direction of the velocity of 20 m s from the top of a multistory resultant of two vectors A and B in terms of their building. The height of the point from where the magnitudes and angle ƒ between them. Fig. 4.9 ball is thrown is 25.0 m from the ground. (a) How (d) A vector A resolved into components along high will the ball rise ? and (b) how long will it be x, y-, and z-axes –1 before the ball hits the ground? Take g = 10 ms–2. Ans. (a) Let us take the y-axis in the vertically upward direction with zero at the ground, as shown in Fig. 3.13. Now 0 = + 20 m s–1, a = – g = –10 m s–2,  = 0 m s–1 If the ball rises to height y from the point of launch, Ans. Let OP and OQ represent the two vectors A and then using the equation 2  20  2 a(y  y0 ) method of vector addition, OS represents the resultant B making an angle . Then, using the parallelogram vector R : R = A + B SN is normal to OP and PM is we get normal to OS. From the geometry of the figure, 0 = (20) + 2(–10)(y – y0) 2 OS2 = ON2 + SN2 Solving, we get, (y – y0) = 20 m. (b) We can solve this part of the problem in two ways. Note carefully but ON = OP + PN = A + B cos  SN = B sin  OS2 = (A + B cos)2 + (B sin )2 the methods used. Ex.5 Rain is falling vertically with a speed of 35 m or, R2 = A2 + B2 + 2AB cos  R  s–1. Winds starts blowing after sometime with a A 2  B2  2AB cos  speed of 12 m s in east to west direction. In which In  OSN, SN = OS sin = R sin, and direction should a boy waiting at a bus stop hold in PSN, SN = PS sin  = B sin  his umbrella ? Therefore, R sin  = B sin  –1 or, R B  sin  sin  (4.24a) (4.24b) Similarly, PM = A sin  = B sin  A B  (4.24c) sin  sin  Combining Eqs. (4.24b) and (4.24c), we get or, Ans. The velocity y of the rain and the wind are represented by the vectors vr and vw in Fig. 4.7 and are in the direction specified by the problem. Using the rule of vector addition, wesee that the resultant R A B   sin  sin  sin  Using Eq. (4.24d), we get: of vr and vw is R as shown in the figure. The magnitude of R is R  r2  2w 2 2  35  12 = m s–1 = 37 m s–1 The direction  that R makes with the vertical is given by tan   sin   w 12   0.343 or, =tan–1 (0.343) = 19º r 35 B sin  R (4.24d) (4.24e) where R is given by Eq. (4.24a). or, tan   SN B sin   OP  PN A  B cos  (4.24f) Equation (4.24a) gives the magnitude of the resultant Therefore, the boy should hold his umbrella in the and Eqs. (4.24e) and (4.24f) its direction. Equation vertical plane at an angle of about 19º with the vertical (4.24a) is known as the Law of cosines and Eq. towards the east. (4.24d) as the Law of sines. KINEMATICS Page # 14 Ex.7 A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat. Ans. The vector vb representing the velocity of the motorboat and the vector vc representing the water current are shown in Fig. 4.11 in directions specified by the problem. Using the parallelogram method of addition, the resultant R is obtained in the direction shown in the figure. We can obtain the magnitude of R using the Law of cosine : R b2  2c  2b c cos 120º  252  102  2  25  10(1 / 2)  22 km / h To obtain the direction, we apply the Law of sines R   c , or, sin  = c sin  sin  sin  R 10  sin120º 10 3   0.397   23.4 21.8 2  21.8 Ex.8 The position of a particle is given by ^ ^ ^ It can be easily verified that the derivative of dr Eq. (4.34a), i.e. gives Eq.(4.33a) and it also dt satisfies the condition that at t=0, r = r0. Equation (4.34a) can be written in component form as 1 x  x0  ax t  ax t2 2 1 ay t2 2 One immediate interpretation of Eq.(4.34b) is that the motions in x- and y-directions can be treated independently of each other. That is, motion in a plane (two-dimensions) can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions. This is an important result and is useful in analysing motion of objects in two dimensions. A similar result holds for three dimensions. The choice of perpendicular directions is convenient in many physical situations, as we shall see in section 4.10 for projectile motion. y  y0  oy t  Ex.9 Rain is falling vertically with a speed of 35 m s–1. A woman rides a bicycle with a speed of 12 m s–1 in east to west direction. What is the direction in which she should hold her umbrella ? Ans. In Fig. 4.16 vr represents the velocity of rain and vb , the velocity of the bicycle, the woman is riding. Both these velocities are with respect to the ground. Since the woman is ridinga bicycle, the velocity of rain as experienced by r  3.0t i  2.0t2 j  5.0 k where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find v(t) and a(t) of the particle. (b) Find the magnitude and direction of v(t) at t = 1.0 s. Ans. v(t)  ^ ^ ^ dr d   3.0t i  2.0t2 j  5.0 k   dt dt   ^ ^ ^ dv   4 .0 j dt along y- direction  3.0 i  4.0 j , a(t)  a = 4.0 m s–2 ^ her is the velocity of rain relative to the velocity of the bicycle she is riding. That is vrb = vr . vb This relative velocity vector as shown in Fig. 4.16 makes an angle ƒ" with the vertical. It is given by ^ At t = 1.0 s, v  3.0 i  4.0 j tan   It's magnitude is = 32  42 = 5.0 m s–1 and direction is  v0 t  1 2 at Or 2 r = r0 + v0t + 1 1 at2 2 Or b 12   0.343 r 35   19º Therefore, the woman should hold her umbrella at an angle of about 19º with the vertical towards the west.