(b)Given the shown v-t diagram (1) construct a-t and s-t diagram (11) Using diagrams, find total distance travelled through the first (35 s). V m mis time s 25s (V4 dingram 6 m/s

For the given velocity time graph

From t = 0 to t = 5 s

velocity  u = 3m/s

Acceleration a = 0 {Velocity is constant}

and the distance traveled

s = v*t = 3 m/s*5 s

   = 15 m

From t = 5 to t = 15 s

velocity  changes from u = 3m/s to u' = -6 m/s

Acceleration a:

by the equation of motion

u' = u+at

-6 m/s = 3 m/s +a (10 s)                                                      (Position curve will be decreasing its slope)

a = -0.9 m/s²

and the distance traveled will be given by the third equation of motion:

$u{\text{'}}^2+u^2 =2as$

$$\begin{gathered} {(-6 m/s)}^2+{(3 m/s)}^2 =2(-0.9 m/s^2)S \\ S= -25 m \end{gathered}$$

So the position of the particle at t = 15 sec will be

s(t =15) = 15 m -25 m

              = -10 m

From t = 15 to t = 25 s

velocity  changes from u' = -6 m/s to v = 5 m/s

Acceleration a:

by the equation of motion

v = u'+at

5 m/s = -6 m/s +a'(10 s)

a = 1.1 m/s²

and the distance traveled will be given by the third equation of motion:

$v^2+u{\text{'}}^2 =2as$

$$\begin{gathered} {(5 m/s)}^2+{(-6 m/s)}^2 =2(1.1 m/s^2)S \\ \mathit{S}\mathbf{\text{'}}\mathbf{=}\mathbf{ }\mathbf{27}\mathbf{.}\mathbf{72}\mathbf{ }\mathit{m} \end{gathered}$$                              (Position curve will be increasing its slope)

So the position of the particle at t = 25 sec will be

s(t = 25) = -10 m +27.72 m

              = 17.72 m