Question

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s = 5 sin πt + 2 cos πt, where t is measured in seconds. (Round your answers to two decimal places.)

(a) Find the average velocity during each time period.
(i) [1, 2]? cm/s
(ii) [1, 1.1]? cm/s
(iii) [1, 1.01]?cm/s
(iv) [1, 1.001]?cm/s
(b) Estimate the instantaneous velocity of the particle when t = 1? cm/s

Answer 1

a.

i. 0 cm/s

ii. 49183 cm/s

iii. 4188.75 cm/s

iv. -673.8 cm/s

b. Instantaneous velocity is -900 cm/s

Step-by-step explanation:

Upon differentiating s = 5 sin πt + 2 cos πt....................Eqn 1

derivative of sin πt is π cos πt

derivative of cos πt is -π sin πt

Therefore derivative of s = 5 sin πt + 2 cos πt is

5π cos πt - 2π sin πt.................Eqn 2

Substituting [1, 2] cm/s into Eqn 2

At t = 1 sec ; 5(180) cos 180 - 2(180) sin 180 = -900 cm/s

At t = 2 sec 5(180) cos 360 - 2(180) sin 360 = 900 cm/s

Average velocity = [900 + (-900)] ÷ 2 = 0

Substituting [1, 1.1] cm/s into Eqn 2

At t = 1 sec ; 5(180) cos 180 - 2(180) sin 180 = -900 cm/s

At t = 1.1 sec 5(180) cos 198 - 2(180) sin 198 = 99,266 cm/s

Average velocity = [99,266cm/s + (-900)] ÷ 2 = 49183 cm/s

Substituting [1, 1.01] cm/s into Eqn 2

At t = 1 sec ; 5(180) cos 180 - 2(180) sin 180 = -900 cm/s

At t = 1.1 sec 5(180) cos 181.8 - 2(180) sin 181.8 = 9277.5 cm/s

Average velocity = [9277.5 + (-900)] ÷ 2 = 4188.75 cm/s

Substituting [1, 1.001] cm/s into Eqn 2

At t = 1 sec ; 5(180) cos 180 - 2(180) sin 180 = -900 cm/s

At t = 1.1 sec 5(180) cos 180.18 - 2(180) sin 180.18 = -447.6 cm/s

Average velocity = [-447.6 + (-900)] ÷ 2 = -673.8 cm/s

b. Instantaneous velocity at t=1

Substituting t=1 into Eqn 2

At t = 1 sec ; 5(180) cos 180 - 2(180) sin 180 = -900 cm/s