Question

The motion of a particle is defined by the relation x = t3 – 6t2 + 9t + 3, where x and t are expressed in feet and seconds, respectively. Determine When the velocity is zero The position, acceleration and total distance traveled when t= 5 sec

Answer 1

Answer:

a. t=3secs and t=1sec

position is -7ft,acceleration is 18fts⁻², total distance travelled is 120ft

Explanation:

the displacement is define as

x=t³-6t²+9t+3·

since we are giving the position as a function of time, the velocity is the derivative of the position,

v=dx/dt

v=d(t³-6t²+9t+3)/dt

recall for y=axⁿ the derivative

dy/dx=a*nxⁿ⁻¹ and the derivative of a constant is zero

hence

V=3t²-12t+9

for V=0,

equivalent to t²-4t+3

solving the quadratic equation, we arrive at

(t-3)(t-1)=0

either t=3 or t=1

hence,at 3secs and 1sec the velocity is zero.

To determine the position at t=5, we substitute t=5 into

t³-6t²+9t+3

(5)³-6(5)²+9(5)+3

125-180+45+3

-7ft

The position at t=5 is -7ft

To determine the acceleration, we differentiate the velocity

a=dv/dt

a=d(3t²-12t+9)/dt

a=6t-12

at t=5

a=6(5)-12

a=18fts⁻²

Next we determine the distance covered at t=5

velocity =total distance travelled/total time taken

velocity=3t²-12t+9

V=3(25)-12(5)+9

V=24ft/s

Hence total distance travelled in t=5 is

24*5=120ft