A particle position as function of time is described as ` y (t) =5 cos (10 t + 15)` in meters. What is the average velocity of the particle from ` t=0` at ` t= 3 sec` ?

Options

A. ` 0. 788 3 m//s`
B. ` - 0. 4314 m//s`
C. ` 0.4313 m//s`
D. `- 0. 7 883 m//s`

Answer

B

Explanation

Here, ` y (t) = 5 cos ( 10 + 15)`
where ` t=0, y (0) = 5 cos ( 10 x 0 + 15) = 5 cos 15^@`
` = 5 xx 0.9659 = 4.8296 m`
when ` t= 3 sec, y (3) = 5 cos ( 10 xx 3 + 15 )`
`= 5 cos 45^@= 5 xx 0. 7081 = 3 .5355 m` ,
total displacement ` =y (3) - y (0)`
` 3. 5355 - 4.8296 =- 1. 29 41 m`
:. ` V_(av)` = (total displacement ) (total time) `= ( -1. 2941)/3`
`=- 0. 4313 m//s`