Question

The position of a particle moving along a straight line is defined by the relation. s = t^3 – 6t^2 – 15t + 40, where s is expressed in feet and t in seconds. Determine:(a) s when t = 3 seconds.(b) v when t = 5 seconds. (c) a when t = 4 seconds.(d) the time when the velocity is equal to zero. What is important about this information?

Answer 1

Answer:

1) s(3) = -32 feet.

2)v(5) = 3 feet/sec

3)a(4) = 12$feet/s^{2}$

4) Velocity becomes zero at t = 5 seconds

Explanation:

Given that position as a function of time is

$s(t)=t^{3}-6t^{2}-15t+40$

Now by definition of velocity we have

$v=\frac{ds}{dt}\\\\v=\frac{d}{dt}(t^{3}-6t^{2}-15t+40)\\\\\therefore v(t)=3t^{2}-12t-15$

Now by definition of acceleration we have

$a=\frac{dv}{dt}\\\\a=\frac{d}{dt}(3t^{2}-12t-15)\\\\\therefore a(t)=6t-12$

Applying values of time in corresponding equations we get

1) s(3)=$3^{3}-6\times (3)^{2}-15\times 3+40=-32feet$

2)v(5)=$3\times {5}^{2}-12\times 5-15=3feet/sec$

3)a(4)=$6\times 4-12=12ft/s^{2}$

4)To obatin the time at which velocity is zero equate the velocity function with zero we get

$3t^{2}-12t-15=0\\\\t^{2}-4t-5=0\\\\t^{2}-5t+t-5=0\\\\t(t-5)+1(t-5)=0\\\\(t-5)(t+1)=0\\\\\therefore t=5\\\\or\\\therefore t=-1$

Thus the correct time is 5 seconds at which velocity becomes zero.