Question

A vehicle is moving at a velocity given by the equation v=5(x^3 +4)/8 find the position at x=4 if when x=2, p=25/2?

Answer 1

Given : A vehicle is moving at a velocity  v = 5(x³ + 4)/8

at x = 2  position p = 25/2

To Find : Position at  x = 4

Solution:

v = 5(x³ + 4)/8

p = ∫vdx

=> p = ∫ 5(x³ + 4)/8 dx

=> p  = (5/8) (x⁴/4  + 4x)  + C

at x = 2  position p = 25/2

=> 25/2  = (5/8)(4 + 8)  + C

=> 25/2 = 60/8  + C

=> 25/2  = 15/2 + C

=> 10/2  = C

=> 5 = C

=> p  = (5/8) (x⁴/4  + 4x)  + 5

at x = 4  

p  = (5/8) (64  + 16)  + 5

=> p  = 55

position at x = 4 is 55

Learn More :

The velocity of particle is given by the equation v=2t²+5. Find the ...

https://brainly.in/question/4711335