A particle moves with a constant acceleration $\left(2 \mathrm{~m} / \mathrm{s}^{2}\right)\( Its initial velocity is \)10 \mathrm{~m} / \mathrm{s}$. Find velocity after \(\mathrm{t}\) second. (A) \((10+\mathrm{t}) \mathrm{ms}^{-1}\) (B) \(5(2+\mathrm{t}) \mathrm{ms}^{-1}\) (C) \(2(5+\mathrm{t}) \mathrm{ms}^{-1}\) (D) \(\left(10+\mathrm{t}^{2}\right) \mathrm{ms}^{-1}\)

Short Answer

Expert verified
The short answer is: \(v = (10 + t) \mathrm{ms}^{-1}\)

Step by step solution

01

Use the velocity formula

To find the velocity after t seconds, apply the formula for the velocity of a particle under constant acceleration: v = u + a*t
02

Plug in the given values

Plug in the given values of initial velocity and constant acceleration: v = 10 m/s + (2 m/s^2) * t
03

Simplify the expression

Simplify the expression by multiplying acceleration (2 m/s^2) with time (t): v = 10 m/s + 2t m/s The particle's velocity after t seconds will be: v = (10 + t) m/s So, the correct answer is: (A) \((10+\mathrm{t}) \mathrm{ms}^{-1}\)

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