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Velocity & Acceleration with Implicit Differentiation

Velocity & Acceleration with Implicit Differentiation. EXAMPLE 1 If 3 s 2 + 5 v 2 = 28 , where s is displacement and v is velocity, find the acceleration a. d (3 s 2 ) + d (5 v 2 ) = d (28) dt dt dt. Substitute. 6 sv + 10 va = 0. 6 s ds + 10 v dv = 0

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Velocity & Acceleration with Implicit Differentiation

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  1. Velocity & Accelerationwith Implicit Differentiation EXAMPLE 1 If 3s2 + 5v2 = 28, where s is displacement and v is velocity, find the acceleration a. d(3s2) + d(5v2) = d(28) dtdtdt Substitute 6sv + 10va = 0 6sds + 10vdv = 0 dt dt 10va = -6sv Remember: ds = v and dv = a dtdt a = – 0.6 s

  2. Assignment Questions 1. The relation between the velocity and distance is given by 5v2 = 40s + 200, where s is the displacement from a fixed point and v is the velocity of a moving object. Find the acceleration. d(5v2) = d(40s) + d(200) dt dt dt 10vdv = 40 ds + 0 dt dt 10va = 40v a = 4 m/s2

  3. Assignment Questions 2. The velocity v of a body is given by the equation 10v2 = 64s + 1000, where s is the distance from a fixed origin. Find the acceleration. d(10v2) = d(64s) + d(1000) dt dt dt 20vdv = 64 ds + 0 dt dt 20va = 64 v a = 64v = 16 = 3.2m/s2 20v 5

  4. Assignment Questions 3. The relation between the velocity vand time tis given by Prove that the acceleration is d(v– 1) + d(⅓)= d(4t) dtdtdt – v -2dv + 0 = 4 dt – 1x a = 4 v2 a = –4v2

  5. Assignment Questions 4. Ifv2 = 4s2 + 200, where v is the velocity of a moving object and s is the distance from a fixed point, find the acceleration at any time. d(v2) = d(4s2) + d(200) dt dt dt 2vdv = 8sds + 0 dt dt 2va = 8sv a = 4 s m/s2

  6. Assignment Questions 5. An object is moving away from a fixed point. The relation between the velocity and the distance is given by 3v2 = 18s + 300, where sis the distance from a fixed point and v is the velocity. Find the acceleration. d(3v2) = d(18s) + d(300) dt dt dt 6vdv = 18 ds + 0 dt dt 6va = 18v a = 3 m/s2

  7. Assignment Questions 6. If s2 + 25v2 = 100 defines a relation between distance s and velocity v, find a relation between acceleration a and distance s. d(s2) + d(25v2) = d(100) dt dt dt 2sds + 50vdv = 0 dt dt 2sv + 50va = 0

  8. EXAMPLE 2 The velocity v of a wind-up toy, in metres per second, is given by v = 300s 3 + s where s is the length of the spring. Find the acceleration in terms of the lengths. a = dv = (3 + s)(300) – 300s(1)ds dt (3 + s)2dt a= 900 + 300s – 300sds (3 + s)2dt ds= v and v = 300s dt 3 + s a = 900 xv (3 + s)2 a = 900 x 300s = 27 000 s (3 + s)23 + s (3 + s)3

  9. Assignment Questions 7. A gun is fired. The bullet has traveled a distance s in the barrel of the gun. The velocity of the bullet is given by v = 6000 s . 6 + s Find the acceleration in terms of s. a = dv = (6 + s)(6000) – 6000s(1)ds dt (6 + s)2dt a = 36 000 + 6000s – 6000sds (6 + s)2dt a = 36 000 v (6 + s)2 a = 36 000 x 6000s = 216 000 000 s (6 + s)26 + s (6 + s)3

  10. Assignment Questions 8. The velocity v is given byv = 1 – 3s2where s is the displacement. 3s2 – 2 Find the acceleration a in terms of s. a = dv = (3s2 – 2)( –6s) – (1 – 3s 2 )(6s)ds dt (3s2 – 2)2dt a = –18s3 + 12s – 6s + 18s3 x 1 – 3s2 (3s2 – 2)23s2 – 2 a = 6 s x 1 – 3s2 (3s2 – 2)23s2 – 2

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