Question

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.

Answer 1

Answer:

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

Answer 2

Answer:

Step-by-step explanation:

hello,

for t = 4 hours

$d(4)=24^3+54^2-3=2*64+5*80-3=128+80-3=205$

this is 205 km

d is differentiable and

$d'(t)=6t^2+10t$

and

$d'(4)=6*4^2+10*4=6*16+40=96+40=136$

this is 136 km/h

d' is differentiable and

$d''(t)=12t+10$

and

$d''(4)=12*4+10=48+10=58$

this is $58 \ km/h^2$