A body moving with an initial velocity U accelerates until it attains a velocity of V within a time t. The distance, s covered by the body is given by the expressions

UNcracked

2 years ago

To find the correct expression for distance \( s \) covered by the body, let's start with the equation of motion:

\[ v = u + at \]

Where:
- \( v \) is the final velocity
- \( u \) is the initial velocity
- \( a \) is the acceleration
- \( t \) is the time taken

Since the body accelerates until it attains velocity \( v \), we know that \( a \) is constant.

Now, using the formula for average velocity:

\[ v_{\text{avg}} = \frac{u + v}{2} \]

We can rearrange to find \( s \):

\[ s = v_{\text{avg}} \times t \]

\[ s = \frac{u + v}{2} \times t \]

\[ s = \frac{(u + u + at)}{2} \times t \]

\[ s = \frac{(2u + at)}{2} \times t \]

\[ s = (u + \frac{1}{2}at) \times t \]

\[ s = ut + \frac{1}{2}at^2 \]

Since \( a = \frac{v - u}{t} \), we can substitute this into the equation:

\[ s = ut + \frac{1}{2} \left( \frac{v - u}{t} \right) t^2 \]

\[ s = ut + \frac{1}{2}(v - u)t \]

\[ s = ut + \frac{1}{2}vt - \frac{1}{2}ut \]

\[ s = \frac{1}{2}vt + \frac{1}{2}ut \]

\[ s = \frac{(v+u)}{2} \times t \]

\[ s = \frac{(v+u)}{2}t \]

So, the correct expression for the distance \( s \) covered by the body is:

\[ \boxed{\text{C. } s = \frac{(v+u)}{2}t} \]

No100

2 years ago

The correct ans should be option A

Since: a=v-u/t
==>S(distance)=1/2at²
=1/2(v-u/t)t²
=(v-u)t/2✅✍️💥