Question

hw3.6. velocity and acceleration of an object
if the velocity of an object in one - dimensional motion is given by (v(t)=-9t^{3}+5t^{2}-8t), where the units of (v) are in m/s and of (t) are in seconds,
the velocity and acceleration of the object at (t = 4.08 s) are:
(a) (v=-561.0 m/s,a = 0 m/s^{2})
(b) (v=-561.0 m/s,a=-417.0 m/s^{2})
(c) (v=-561.0 m/s,a=-834.0 m/s^{2})
(d) (v=-561.0 m/s,a = 9.81 m/s^{2})
(e) (v = 4.08 m/s,a=-209.0 m/s^{2})
(f) (v=-561.0 m/s,a = 417.0 m/s^{2})

Explanation:

Step1: Calculate velocity at $t = 4.08$ s

Substitute $t = 4.08$ into $v(t)=-9t^{3}+5t^{2}-8t$.
$v(4.08)=-9\times(4.08)^{3}+5\times(4.08)^{2}-8\times4.08$
$v(4.08)=-9\times67.917312 + 5\times16.6464-32.64$
$v(4.08)=-611.255808+83.232 - 32.64\approx - 561.0$ m/s

Step2: Find acceleration function

Acceleration $a(t)$ is the derivative of velocity function $v(t)$.
$v(t)=-9t^{3}+5t^{2}-8t$, so $a(t)=v^\prime(t)=-27t^{2}+10t - 8$.

Step3: Calculate acceleration at $t = 4.08$ s

Substitute $t = 4.08$ into $a(t)=-27t^{2}+10t - 8$.
$a(4.08)=-27\times(4.08)^{2}+10\times4.08-8$
$a(4.08)=-27\times16.6464 + 40.8-8$
$a(4.08)=-449.4528+40.8 - 8\approx - 417.0$ m/s²

Answer:

B. $v=-561.0$ m/s, $a = - 417.0$ m/s²