Question

An object is at x=0 at t=0 and moves along the xaxis according to the velocity-time graph in Figure P2.62.

(a) What is the object's acceleration between 0 and 4.0s?

(b) What is the object's acceleration between 4.0sand 9.0s?

(c) What is the object's acceleration between 13.0s and 18.0s?

(d) At what time(s) is the object moving with the lowest speed?

(e) At what time is the object farthest from x=0?

(f) What is the final position x of the object at t=18.0s?

(g) Through what total distance has the object moved between t=0 and t=18.0s?

Short answer

$$\begin{gathered} a)Theobject\text{'}saccelerationbetween0and4.0sis0. \\ b)Theobject\text{'}saccelerationbetween4.0sand9.0sis6.0m/s^2. \\ c)Theobject\text{'}saccelerationbetween13.0sand18.0sis-3.6m. \\ d)Thetime\left(s\right)oftheobjectmovingwiththelowestspeedist=6sandt=18s. \\ e)Thetimeistheobjectfarthestfromx=0ist=18s. \\ f)Thefinalpositionxoftheobjectatt=18.0sis84m. \\ g)totaldis\tan cehastheobjectmovedbetweent=0andt=18.0sis204m. \end{gathered}$$

Step-by-step solution

Definition of velocity

The rate of change of an object's location with regard to a frame of reference is its velocity, which is a function of time. A statement of an object's speed and direction of motion is referred to as velocity.

Finding the object's acceleration between 0 and 4.0s.

(a)

The acceleration is zero between t=0 and t=4s since the velocity is constant.

Finding the object's acceleration between 4.0s and 9.0s

$$\begin{gathered} \left(b\right) \\ Theratioofthechangeinvelocity∆v,dividedbythetimeinterval∆tduringwhich \\ thatchangeoccurs,istheaverageacceleration. \\ a=\frac{∆v}{∆t} \\ =\frac{v_9-v_4}{9-4} \\ =6m/s^2 \end{gathered}$$

Finding the object's acceleration between 13.0s and 18.0s

$$\begin{gathered} \left(c\right) \\ Theaccelerationoftheobject, \\ a=\frac{∆v}{∆t} \\ =\frac{v_{18}-v_{13}}{18-13} \\ =-3.6m/s^2 \end{gathered}$$

Finding the time(s) is the object moving with the lowest speed

(d)

In the graph, the speed is zero at t=6s and t=18s .

Finding the time is the object farthest from x=0

(e)

In the graph, the item reaches its maximum distance at t=18s .

Finding the final position x of the object at t=18.0s

$$\begin{gathered} \left(f\right) \\ Theentireareabeneaththegraphtoobtainthefinallocationatt=18m. \\ Thegraphisdividedintofiveareas(tworec\tan glesandthreetriangles), \\ Areaofatriangle=\frac{1}{2}\left(base\right)\left(height\right) \\ Areaofarec\tan gle=length\times width \\ ∆x=\left(-12\right)\left(4\right)+\frac{1}{2}\left(-12\right)\left(2\right)+\frac{1}{2}\left(18\right)\left(3\right)+\left(18\right)\left(4\right)+\frac{1}{2}\left(18\right)\left(5\right) \\ =84m \end{gathered}$$

Finding the total distance has the object moved between t=0 and t=18.0s

$$\begin{gathered} \left(g\right) \\ We\text{'}llusetheabsolutevaluesofthevariablesinparttocomputethetotaldis\tan ce \\ ratherthanthedisplacement\left(f\right),). \\ d=\left(12\right)\left(4\right)+\frac{1}{2}\left(12\right)\left(2\right)+\frac{1}{2}\left(18\right)\left(3\right)+\left(18\right)\left(4\right)+\frac{1}{2}\left(18\right)\left(5\right) \\ =204m \end{gathered}$$