Calculus Question

Question

Theres three parts to this question

If x(t)=−2t^2+6t+3x(t).

find the position of the object at t=3.7s.t=3.7s. Use standard SI units.

Find the velocity of the object at t=3.7s

Find the accelration of the object at t=3.7s

Answer 1

x(t) =-2x^2 +6x + 3

x(3.7) =-2(3.7)^2 +6(-3.7) +3

= -2(13.69) -22.2 +3

= -27.38 - 19.2

=-46.58 meters

the object moved backwards to a point -46.58 meters

it moved backwards 46.58-3 = 43.58 meters from its initial position

velocity is the derivative of the distance function

v(t)=x'(t) = -4x +6 = -4(3.7) +6

= -14.8+6

=- 8.8 meters per second

velocity = -8.8 m/s when time t=3.7 seconds

acceleration = the derivative of the velocity function = the 2nd derivative of the distance function

a(t) = v'(t) = s"(t)

a(t) = -4 meters per second per second

acceleration = a constant -4 m/s^2

the velocity is slowing down regardless of the time