Question

please answer it is urgent i will mark brainliest​

Answer 1

Answer:

Your answer is D..

acceleration is constant and value of acceleration is 2 m/s^2

Answer 2

Answer:

$\sqrt{x} = t + 9 \\ \\ squaring \: both \: side \\ = > ( { \sqrt{x})}^{2} = {(t + 9)}^{2} \\ = > x \: = {t}^{2} + 2 \times t \times 9 + {9}^{2} \\ = > x = {t}^{2} + 18t + 81 \\ as \: we \: know \: that \: \\ \\ velocity(v) = \frac{dx}{dt} \\ = > v = \frac{d( {t}^{2} + 18t + 81)}{dt} \\ = > v = 2 {t}^{2 - 1} + 18 \times 1 \times {t}^{1 - 1} + 0 \\ = > v =( 2t + 18) \: \: \frac{m}{s} \\ \\ also \\ \\ acceleration(a) = \frac{dv}{dt} \\ = > a = \frac{d(2t + 8)}{dt} \\ = > a = 2 \times 1 \times {t}^{1 - 1} + 0 \\ = > a= 2 \: {t}^{0} \: \: \: \: | {t}^{0} = 1 | \\ = > a = 2 \: \frac{m}{ {s}^{2} }$

Explanation:

• It is clear from the above results,
• Velocity is directly propotional to the time(t)
• Acceleration is constant I. e, it does not depend on the time

Hence, correct answer will be (d)