Question

Assuming the acceleration in example 2 above remains constant, how long will Mahendra take to move 1 cm towards Virat?​

Answer 1

Answer:

According to the information in Example 2,

Initial Velocity (u) = 0 m/s

Acceleration = 5.34 × 10⁻⁹ m/s²

Distance to be covered = 1 cm (or) 10⁻² m.

Time taken = ?

Using the second equation of motion we get:

⇒ s = ut + 0.5 at²

⇒ 10⁻² = (0 × t) + (0.5 × 5.34 × 10⁻⁹ × t² )

⇒ 10⁻² = 2.67 × 10⁻⁹ × t²

$\implies t^2 = \dfrac{10^{-2}}{2.67 \times 10^{-9}}\\\\\\\implies t^2 = \dfrac{1}{2.67} \times 10^{(-2+9)}\\\\\\\implies t^2 = 0.37 \times 10^7 = 3.7 \times 10^6\\\\\ \implies t = \sqrt{3.7 \times 10^6}\\\\ \implies t = 1.9235 \times 10^3\:\:sec\\\\ \implies \boxed{ \bf{t = 1923.5\:\:seconds}}$

Hence it would take 1923.5 seconds to cover 1 cm when the acceleration is constant.

Answer 2

Explanation:

...its ans in photo ....