Engineering Physics

Question

A particle moving along the -axis has its position described by the function , where x=(5t^3+2t+3) where t is in s

Part A- At t=2s what is the particle's position?

Express your answer to three significant figures and include the appropriate units. .

Part -B At t=2s, what is the particle's velocity?

Express your answer to three significant figures and include the appropriate units. .

Part- C At =2, what is the particle's acceleration?

Answer 1

Part A X = 5t³ +2t +3

At t=2 S

Position X = 5 (2)³ + 2*2 +3 = 40+4 + 3= 47 m or units as in question for units of X

Part B

Speed = dX/dt

so derivative of X wrt t

dX/dt = d/dt ( 5t³ +2t +3) = V

= 15 t² +2t +0

Plug in t = 2 sec

Part C

Acceleration = d/dt ( V)

d/dt( 15t2 +2t) = a

= 15*2 t +2 unit m/s

=30t +2 = 62 m/s² ( For t= 2 Sec)

Answer 2

(A)

Evaluate x(2). Just plug t=2 into the equation for x(t).

(B)

v(t) = dx/dt = 15t² + 2

Evaluate v(2).

(C)

a(t) = dv/dt = 30t

Evaluate a(2).