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The velocity vs. time graph for an object moving along a straight path is shown in the figure below.(m/s)Find the average acceleration of this object during the following time intervals.(a) 0 s to 5.0 sm/s²(b) 5.0 s to 15 sm/s²(C) 0 s to 20 sm/s²(ii) Find the instantaneous accelerations at the following times.(a) 2.0 sm/s²(b) 10 sm/s²(c) 18 sm/s²

The Velocity Vs Time Graph For An Object Moving Along A Straight Path Is Shown In The Figure BelowmsFind The Average Acceleration Of This Object During The Foll class=

Sagot :

ANSWER:

(i)

(a) 0 m/s²

(b) 1.6 m/s²

(c) 0.8 m/s²

(ii)

(a) 0 m/s²

(b) 1.6 m/s²

(c) 0 m/s²

STEP-BY-STEP EXPLANATION:

(i)

We have that the acceleration is given as follows:

[tex]a=\frac{\Delta v}{\Delta t}[/tex]

(a) 0 s to 5 s:

[tex]a=\frac{v_5-v_0}{5-0}=\frac{-8-(-8)}{5-0}=\frac{-8+8}{5}=\frac{0}{5}=0\text{ }\frac{m}{s^2}[/tex]

(b) 5 s to 15 s:

[tex]a=\frac{v_{15}-v_5}{15-5}=\frac{8-(-8)}{15-5}=\frac{8+8}{10}=\frac{16}{10}=1.6\text{ }\frac{m}{s^2}[/tex]

(c) 0 s to 20 s:

[tex]a=\frac{v_{20}-v_0}{20-0}=\frac{8-(-8)}{20-0}=\frac{8+8}{20}=\frac{16}{20}=0.8\text{ }\frac{m}{s^2}[/tex]

(ii)

The instantaneous acceleration is given as follows:

[tex]a=\frac{dv}{\text{ dt}}\text{ (slope of graph)}[/tex]

(a) 2 sec (No slope so acceleration at 2 sec is 0 m/s²). We confirm it as follows:

[tex]a=\frac{v_2-v_0}{2-0}=\frac{-8-(-8)}{2-0}=\frac{-8+8}{2}=\frac{0}{2}=0\text{ }\frac{m}{s^2}[/tex]

(b) 10 sec

The slope is:

[tex]a=\frac{8-(-8)}{15-5}=\frac{8+8}{10}=\frac{16}{10}=1.6\text{ }\frac{m}{s^2}[/tex]

(c) 18 sec: No slope so acceleration at 18 sec is 0 m/s²

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