Question

the velocity time graph represent motion of a cyclist along straight line find the total distance travelled acceleration retardation average velocity​

Answer 1

$\huge{\bold{\underline{\underline{Solution:-}}}}$

$\large{\bold{\underline{To Find}}}$

• Distance travelled =?.
• Acceleration = ?.
• Retardation = ?.
• Average velocity = ?.

$\large{\bold{\underline{\underline{Step - by - Step - Explanation}}}}$

• Case-1

Distance =?.

As you know the area under velocity time graph gives distance.

So,

${S = ar( \triangle)AOD + ar( \triangle)BEC + ar( \square)ABDE}$

${S = \frac{1}{2} \times AD \times OD + \frac{1}{2} \times BE \times EC + DE \times BE}$

${ \therefore S = \frac{1}{2} \times 40 \times 5 + \frac{1}{2} \times 40 \times 2 + 2 \times 40}$

${S = 20 \times 5 + 20 \times 2 + 40 \times 2}$

${S = 100 + 40 + 80}$

$\huge{\boxed{\boxed{S = 220 \: metres}}}$

• Case-2

Acceleration = ?.

Slope of velocity time graph gives acceleration.

so,

$\huge{a = \frac{y_2 - y_1}{x_2 - x_1}}$

${ \therefore a = \frac{ 40 - 0}{5 -0}}$

${ \therefore a = \frac{40}{5}}$

$\huge{\boxed{\boxed{a = 8 m/s^2}}}$

• Case-3

Retardation = ?.

Slope of velocity time graph gives retardation.

so,

$\huge{r = \frac{y_2 - y_1}{x_2 - x_1}}$

${ \therefore r = \frac{ 0 - 40}{11 -7}}$

${ \therefore r = \frac{-40}{2}}$

$\huge{\boxed{\boxed{r = - 20 m/s^2}}}$

• Case-4

Average velocity = ?.

${Average \: velocity = \frac{Total \: distance}{Total \: Time}}$

${ \therefore v_{av} = \frac{220}{11}}$

$\huge{\boxed{\boxed{v_{av} = 20 m/s}}}$