Please Help Me The displacement of a particle d (in km) as a function of time t (in hours) is given by: d(t) = 2t^3 + 5t^2 − 3. Find the displacement, velocity and acceleration at t = 4 hours. Indicate the correct units for each of these quantities.

Step-by-step explanation:

hello,

for t = 4 hours

this is 205 km

d is differentiable and

and

this is 136 km/h

d' is differentiable and

and

this is

Step-by-step explanation:

Displacement:  d(t) = 2t^3 + 5t^2 - 3

Velocity:  (d/dt)(2t^3 + 5t^2 - 3) = v(t) = 6t^2 + 10t

Acceleration:  (d/dt)(d/dt)(6t^2 + 10t) = a(t) = 12t + 10

At time t = 4hr,

d(4) = [128 + 80 - 3] km = 205 km

v(4) = [96 + 40] km/hr  =  136 km/hr

a(4) = [48 + 10] km per hour squared  = 58 km over hours squared

step-by-step explan mxfcg'k; uafwgzfgjhklhgation:

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!