A particle is moving along a straight line with constant acceleration. At the end of tenth second its velocity becomes 20 m/s and in tenth second it travels a distance of 10 m. Then the acceleration of the particle will be

Question

A particle is moving along a straight line with constant acceleration. At the end of tenth second its velocity becomes 20 m/s and in tenth second it travels a distance of 10 m. Then the acceleration of the particle will be (A) 10 m/s2 (B) 20 m/s2 (C) (1/5) m/s2 (D) 3.8 m/s2

Tardigrade Admin · Accepted Answer

Using v - u + at ⇒ 20 = u + a × 10 ...(i) and sn = u + (a/2) ( 2n - 1) ⇒ 10 = u + (a/2) ( 2× 10 - 1) ...(ii) On solving (i) and (ii) we get a = 20 m/s2

Q. A particle is moving along a straight line with constant acceleration. At the end of tenth second its velocity becomes $20 m / s$ and in tenth second it travels a distance of $10 m$ . Then the acceleration of the particle will be

$10 m / s^{2}$

$20 m / s^{2}$

$\frac{1}{5} m / s^{2}$

$3.8 m / s^{2}$

Using $v - u + a t$

$\Rightarrow 20 = u + a \times 10 ... (i)$

and $s_{n} = u + \frac{a}{2} (2 n - 1)$

$\Rightarrow 10 = u + \frac{a}{2} (2 \times 10 - 1) ... (ii)$

On solving (i) and (ii) we get $a = 20 m / s^{2}$

Q. A particle is moving along a straight line with constant acceleration. At the end of tenth second its velocity becomes 20m/s and in tenth second it travels a distance of 10m. Then the acceleration of the particle will be

10m/s2

20m/s2

51​m/s2

3.8m/s2