Question

If the velocity 'u' of a body moving along a straight line with a constant retardation 'a' is reduced by
75% in time t', then t-
options in the ppic sorry physics

Answer 1

Answer:-

Your Answer Is Option b) $\bf\dfrac{3u}{4a}.$

Explanation:-

Given:-

• Initial Velocity = u.

• Final velocity = reduced by 75% from the Initial Velocity.

• Constant Retardation (so the acceleration would be negative) = a.

• So Acceleration = -a.

• Time = t.

To Find:-

• The expression for t.

Equation Used:-

$\bf \ratio\implies v = u + at.$

Where,

• v = Final Velocity.
• u = Initial Velocity.
• a = Constant Acceleration.
• t = Time.

So Here,

• $\tt v = u - 75\%\:\: of \:\:u.$
• u = u.
• a = (-a).
• t = t.

Solution:-

As in the que. it is given that the initial velocity "u" ir reduced by 75% by acceleration "-a" in time "t".

So Let the Final Velocity be v.

$\\ \tt\mapsto v = u - 75 \% \: \: \: of \: \: \: u.$

$\\ \tt\mapsto v = u - \frac{75}{100} \times u.$

$\\ \tt\mapsto v = u - \frac{75u}{100} .$

$\\ \tt\mapsto v = \frac{100u - 75u}{100} .$

$\\ \tt\mapsto v = \frac{ \cancel{25} \: \: u}{ \cancel{100}}.$

$\\ \large\bf\mapsto v = \frac{u}{4} ..... \: eq(1).$

By Using Equation we get:-

$\\ \tt\mapsto v = u + at.$

$\\ \tt\mapsto v = u + ( - a) \times t.$

$\\ \bf \large\mapsto v = u - at.....eq(2).$

So From eq(1) and eq(2) we get:-

$\\ \tt\mapsto u - at = \dfrac{u}{4}.$

$\\ \tt\mapsto - at = \frac{u}{4} - u.$

$\\ \tt\mapsto - at = \frac{ - 3u}{4} .$

$\\ \tt\mapsto t = \frac{ \cancel{ - } \: \: 3u}{ \cancel{ - } \: \: 4a} .$

$\\ \bf \large\mapsto{ \underline{ \boxed{ \bf t = \frac{3u}{4a}. }}}$

$\bf\therefore$ The Final Answer Is Option b.

Answer 2

Aɴsᴡᴇʀ :

$\:$

• Time (t) = $\boxed{\tt{\red{\dfrac{3u}{4a}}}}$

$\:$

Gɪᴠᴇɴ :

$\:$

• Initial velocity = u

• Acceleration = -a (retardation)

• Time = t

• Final velocity (v) = u - 75% of u

$\:$

Tᴏ Fɪɴᴅ :

$\:$

• Time in terms of u and a.

$\:$

Fᴏʀᴍᴜʟᴀ Usᴇᴅ :

$\:$

• $\boxed{\sf{v=u+at}}$

$\:$

Sᴏʟᴜᴛɪᴏɴ :

$\:$

➪ $\sf{v = u - 75\% \:of \:u}$

$\:$

➪ $\sf{v = 25\% \:of\: u}$

$\:$

➪ $\sf{v = u\times\dfrac{25}{100}}$

$\:$

➪ $\sf{v = \dfrac{u}{4}}$ --(i)

$\:$

By substituting (i) in formula, we get

$\:$

➪ $\sf{v=u+at}$

$\:$

➪ $\sf{\dfrac{u}{4}=u+at}$

$\:$

Here ‘a’ is retardation (-a), so

$\:$

➪ $\sf{\dfrac{u}{4} = u-at}$

$\:$

➪ $\sf{at = \dfrac{3u}{4}}$

$\:$

➪ $\sf{t = \dfrac{3u}{4a}}$

$\:$

★ Hence, option b is correct ✔