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Velocity & Acceleration with Implicit DifferentiationPowerPoint Presentation

Velocity & Acceleration with Implicit Differentiation

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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

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

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

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

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

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

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

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

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

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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