Question

A) Object starts moving from the rest with uniform acceleration and travels 20 m in first 2 s what will be it’s velocity after 10 s from the start and displacement of particle in 10 th seconds? B ). An object starts from rest with uniform acceleration and covers 100m in 10s. Find speed of object just after covering 50 m

Answer 1

Answer:

A) 100 m/s, 500 m

B) $10\sqrt{2}$ m/s

Explanation:

Formulae:

$1. v =u+at$

$2. s = ut+\dfrac{1}{2}at^{2}$

$3. v^{2} -u^{2} = 2as$

Part (A)

Given :

Initial velocity, $u = 0\ m/s$

Displacement, $s = 20 m$

Time, $t = 2s$

Let us find acceleration, $a$ Using Formula 2:

$20 = 0 \times 2 + \dfrac{1}{2} \times a \times 2^{2} \\\Rightarrow 20 = \dfrac{1}{2} \times a \times 4\\\Rightarrow 2a = 20\\\Rightarrow a = 10 m/s^{2}$

Using Formula (1) to find Velocity after 10s from start:

$v = 0 + 10(10)\\\Rightarrow v = 100 m/s$

Using formula (2) to find displacement of particle in 10 seconds:

$s = 0 \times 10 + \dfrac{1}{2} \times 10 \times 10^{2} \\\Rightarrow s = 5 \times 100\\\Rightarrow s = 500 m$

Part (B)

Given:

Initial velocity, $u = 0\ m/s$

Displacement, $s = 100 m$

Time, $t = 10s$

Let us find acceleration, $a$ Using Formula 2:

$100 = 0 \times 10 + \dfrac{1}{2} \times a \times 10^{2} \\\Rightarrow 100 = \dfrac{1}{2} \times a \times 10^2\\\Rightarrow 50a = 100\\\Rightarrow a = 2 m/s^{2}$

Using Formula (3) to find speed of object, $v$ just after covering distance, $s = 50 m$

$v^{2} -0^{2} = 2 \times 2 \times 50\\\Rightarrow v^2 = 200\\\Rightarrow v = \sqrt{200}\\\Rightarrow v = 10\sqrt{2}\ m/s$