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

an altitude of 15,000 feet

step-by-step explanation:

answer

a,b, and d

explanation

a reflection angle is more than

but less than;

an angle that measures

is a reflex angle.(true)

an angle that measures less than

is acute.(true)

provided we are considering angles greater or equal to zero.

an acute angle lies in the range.

an obtuse angle is less than 180° but more than 90°

an angle that measures 300° is obtuse (false)

an angle that measures 65° is an acute angle.(true)