What is the initial velocity of the particle that covers 46 m in the 9th and 10th seconds?

Steps to Solve

1. Understanding Relevant Variables

We will use the formula for displacement in the nth second of motion for uniformly accelerated motion, which is given by: $$ s_n = u + \frac{1}{2} a (2n - 1) $$

Here,

• $s_n$ is the distance covered in the nth second,
• $u$ is the initial velocity,
• $a$ is the acceleration, and
• $n$ is the second we are interested in.

2. Calculate Distance for 9th and 10th Seconds

We want to calculate the distance covered during the 9th second ($s_9$) and the 10th second ($s_{10}$).

Using the formula: $$ s_9 = u + \frac{1}{2} a (2 \cdot 9 - 1) = u + \frac{1}{2} a (17) = u + \frac{17}{2} a $$

$$ s_{10} = u + \frac{1}{2} a (2 \cdot 10 - 1) = u + \frac{1}{2} a (19) = u + \frac{19}{2} a $$

3. Find Total Distance Over 9th and 10th Seconds

To find the total distance covered during the 9th and 10th seconds, we add $s_9$ and $s_{10}$: $$ s_9 + s_{10} = \left( u + \frac{17}{2} a \right) + \left( u + \frac{19}{2} a \right) $$ Combine like terms: $$ s_9 + s_{10} = 2u + 18a $$

4. Solve for Initial Velocity

If we know the total distance $D$ covered during the 9th and 10th seconds, we can rearrange the equation to solve for $u$: $$ 2u + 18a = D $$ Finally, isolate $u$: $$ u = \frac{D - 18a}{2} $$

5. Substituting Known Values

Plug in the known values for $D$ and $a$ into the equation to compute for $u$.