Question

Answer it............​

Answer 1

★ Answer :

(i) [0 - 10 s]

→ Distance = Area of OAE

$\rm{\implies x_1 = \frac{1}{2} \times Base \times Height} \\ \\ \rm{\implies x_1 = \frac{1}{2} \times OE \times AE} \\ \\ \rm{\implies x_1 = \frac{1}{2} \times 10 \times 10} \\ \\ \rm{\implies x_1 = \frac{\cancel{100}}{\cancel{2}}} \\ \\ \rm{\implies x_1 = Displacement = 50 \: cm} \\ \\ \Large{\star{\boxed{\tt{x_1 = 50 \: cm}}}}$

$\rule{200}{2}$

(ii) [10 - 30 s]

$\rm{x_2 = Distance = Area \: of \: trapezium \: EABC} \\ \\ \rm{\implies x_2 = \frac{1}{2} (Sum \: of \: parallel \: sides) \times Height} \\ \\ \rm{\implies x_2 = \frac{1}{2} (AB + EC) \times AE} \\ \\ \rm{\implies x_2 = \frac{1}{2} (10 + 20) \times 10} \\ \\ \rm{\implies x_2 = \frac{\cancel{300}}{\cancel{2}}} \\ \\ \rm{\implies x_2 = 150\: m} \\ \\ \Large{\star{\boxed{\tt{x_2 = 150 \: m}}}}$

Answer 2

Answer:

Refer to attachment please.

Here, it is not specified that it is an velocity-time curve or speed-time curve,so we can say that area under the curve along time axis will give both distance and displacement.

Thus, distance=displacement.