Question

Figure (3-E4) shows the graph of the x-coordinate of a particle going along the X-axis as a function of time. Find (a) the average velocity during 0 to 10 s, (b) instantaneous velocity at 2, 5, 9 and 12s.
Figure

Answer 1

Answer:

Figure (3-E4) shows the graph of the x-coordinate of a particle going along the X-axis as a function of time. Find  (a) the average velocity during 0 to 10 s,  (b) instantaneous velocity at 2, 5, 8 and 12s.

Explanation:

(a) Displacement during 0 to 10 s = x = 100 m,    Average velocity during 0 to 10 s = 100 m/ 10 s = 10 m/s 

(b) Instantaneous velocity at n second,  vn is given by the slope of the graph at that instant   

v2 = slope of st line at 2s = 50/2.5 = 20 m/s  

v5 = slope of st line at 5 s = zero    

v8 = slope of st line at 8 s = (100-50)/(10-7.5) =50/2.5 =20 m/s  

v12 = slope of st line at 12 s = -100/5 =-20 m/s  (because slope is negative)

Answer 2

(a) Average velocity during 0 to 10 s = 10 m/s.

(b) Instantaneous velocity during at 2, 5, 8 and 12 s : $v_2$ = 20 m/s    ;     $v_5$ = 0    ;    $v_8$ = 20 m/s     ;    $v_1_2$ = - 20 m/s

Explanation:

(a) Average velocity during 0 to 10 s :

Displacement 0 to 10 second  = x = 100 m,    

Average velocity during 0 to 10 s  $=\frac{\text {displacement}}{\text {time}}$

                                                         $=\frac{100 m}{10 s}$

                                                         =  10 m/s  

(b) Instantaneous velocity during at 2, 5, 8 and 12 s :

Instant velocity at  "n" second

At that moment, $v_n$ is given by the graph slope

$\mathrm{v}_{2}=\text { slope of st line at } 2 \mathrm{s}=\frac{50}{2.5}=20 \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{5}=\text { slope of st line at } 5 \mathrm{s}=\mathrm{zero}$

$\mathrm{v}_{8}=\text { slope of st line at } 8 \mathrm{s}=\frac{100-50}{10-7.5}=\frac{50}{2.5}=20 \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{12}=\text { slope of st line at } 12 \mathrm{s}=\frac{-100}{5}=-20 \mathrm{m} / \mathrm{s}$