Given () = −^3 + 2^2 +(3/2) be the position of a particle moving along the x-axis at time t. At what time will the instantaneous velocity equal the average velocity over the time interval [0, 4]

5.266 secs

Step-by-step explanation:

Lets assume ; p(t) = t^-3 + 2^2 + ( 3/2 ) is the particle position along x-axis

time interval [ 0, 4 ]

Average velocity = Displacement / time

                            = p( b ) - p( a ) /  b - a -------- ( 1 )

where a = 0 , b = 4 ( time intervals )

Back to equation 1

Average velocity = [ ( 4^-3 + 4 + (3/2) ) - ( 0 +  4 + (3/2) ) ] / 4

                             = 3.9 * 10^-3 ----- ( 2 )

Instantaneous velocity = d/dx p(t)

                                      = - 3/t^4  ------ ( 3 )

To determine the time that the instanteous velocity = average velocity

equate equations (2) and (3)

3.9*10^-3 = - 3 / t^4

t^4 = - 3 / ( 3.9 * 10^-3 ) = - 769.231

hence t = = 5.266 secs

we ignore the negative sign because time can not be in the negative

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol

p(x)= 8 x³ - 36 x² + 54 x - 27

q(x) = 2 x -3

you can check your mistake and then you can choose the correct option.

dividend =   8 x³ - 36 x² + 54 x - 27

divisor = 2 x -3

if you want to find quotient, proceed as follows

q(x)= 0

2 x - 3=0

x =

now , i have used division method , to find quotient and remainder