Question

a 7. the displacement x of an object moving along the x - axis is shown above as a function of time t. the acceleration of this object must be (a) zero (b) constant but not zero (c) increasing (d) decreasing (e) equal to g c 8. an equation for the motion of an object is in the mks system, and is described by the function: x(t)=7.5t³ + 9t + 18 then the dimensions of the coefficient “7.5” must be (a) l/t (b) l/t² (c) l/t³ (d) t²/l (e) t³ e 9. which of the following would be a correct equation for the velocity with respect to time for this object? (a) v(t)=7.5t + 9 (b) v(t)=7.5t² + 9 (c) v(t)=22.5t + 9 (d) v(t)=22.5t³ + 9t (e) v(t)=22.5t² + 9 c 10. a body moving in the positive x direction passes the origin at time t = 0. between t = 0 and t = 1 second, the body has a constant speed of 24 meters per second. at t = 1 second, the body is given a constant acceleration of 6 meters per second squared in the negative x direction. the position x of the body at t = 11 seconds is (a) +99m (b) +36m (c) - 36 m (d) - 75 m (e) - 99 m questions 11 - 12. the object shown in the position - time graph above. (a) started off slowly, sped up, and then slowed down again. (b) started off rapidly, slowed down, and then sped up again. (c) has a constant positive acceleration. (d) has a constant negative acceleration. (e) both (b) and (d) the magnitude of the acceleration of the object is about (a) 0.56 m/s/s (b) 1.2 m/s/s (c) 3 m/s/s (d) 9.8 m/s/s (e) cannot be determined

Explanation:

Step1: Recall velocity - time relationship

Velocity $v(t)$ is the derivative of position $x(t)$. Given $x(t)=7.5t^{3}+9t + 18$, using the power - rule of differentiation $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=\frac{d}{dt}(7.5t^{3}+9t + 18)$.

Step2: Differentiate each term

$\frac{d}{dt}(7.5t^{3})=7.5\times3t^{2}=22.5t^{2}$, $\frac{d}{dt}(9t)=9$, and $\frac{d}{dt}(18)=0$. So $v(t)=22.5t^{2}+9$.

Answer:

E. $v(t)=22.5t^{2}+9$