Question

a body starting from the rest moves with a uniform acceleration of 0.02m/s from 5minutes calculate the speed acquired and the dista travelled in this time

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Answer 1

$\large \dag$ Question :-

A body starting from the rest moves with a uniform acceleration of 0.02 m/s² for 5 minutes. Calculate the speed acquired and the distance travelled in this time.

$\large \dag$ Answer :-

${\begin{cases} \bf Speed\:Acquired = 6 \:m/s \\  \\  \bf Distance\:Travelled = 900 \: m\end{cases}}$

$\large \dag$ Step by step Explanation :-

❒ Converting Time in Seconds :-

We Have,

$\text{Time= 5 mins}$

$= \rm( 5 \times 60)second$

$\red{:\longmapsto \text{Time = 300 \:s}}$

Now Here we have :

• Initial Velocity = u = 0 m/s

• Acceleration = a = 0.02 m/s²

• Time = t = 300 s

1 》 Calculating Speed Acquired :-)

Let Speed Acquired by Body be = v m/s

❒ We Have 1st Equation of Motion as :

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{v = u + at}}}$

⏩ Applying 1st Equation of Motion ;

$:\longmapsto \rm v = 0 + 0.02 \times 300$

$:\longmapsto \rm v = 0.02 \times 300$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf v = 6 \: m/s} }}}$

Therefore,

$\large\underline{\pink{\underline{\frak{\pmb{\text Speed \: \text Acquired = 6\: m/s }}}}}$

2 》Calculating Distance Traveled :-)

❒ We Have 2nd Equation of Motion as :

$\large \bf \red \bigstar \: \: \orange{ \underbrace{ \underline{s=ut+\dfrac{1}{2}at^2}}}$

⏩ Applying 2nd Equation of Motion :

$:\longmapsto \rm s = 0 \times 300 + \frac{1}{2} \times 0.02 \times {300}^{2}$

$:\longmapsto \rm 0 + \frac{1}{2} \times 0.02 \times {300}^{2}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf s = 900 \: m} }}}$

Therefore,

$\large\underline{\pink{\underline{\frak{\pmb{Distance \: \text Traveled = 900 \: m }}}}}$

Answer 2

Answer:

‣ Speed acquired $\mapsto\:{\boxed{\tt{6\:m/s}}}$

‣ travelled $\mapsto\:{\boxed{\tt{900\:m}}}$

Explanation:

Given information,

A body starting from the rest moves with a uniform acceleration of 0.02 m/s² from 5 minutes calculate the speed acquired and the distance travelled in this time.

We have to find speed acquired i.e, final velocity (v) and distance travelled (s);

We have,

• Initial velocity (u) = 0 m/s
• Acceleration (a) = 0.02 m/s²
• Time taken (t) = 5 minutes
• Final velocity (v) = ?
• Distance travelled (s) = ?

★ Converting units of time ::

⚘ [1 minitue = 60 seconds]

• Time in seconds = (5 × 60) s
• Time in seconds = 300 s

★ Finding final velocity (v) ::

⚘ Using first eqⁿ of motion;

• $\boxed{\underline{\underline{\bf{\red{v = u + at}}}}}\:\bf{\dag}$

Putting all values,

$\\ \longrightarrow\:\tt v = 0 + (0.02)(300)$

$\\ \longrightarrow\:\tt v = 0 + (0.02\:\times\:300)$

$\\ \longrightarrow\:\tt v = 0 + 6.00$

$\\ \longrightarrow\:{\underline{\boxed{\tt{v = 6\:m/s}}}}\:\bigstar$

• Hence, speed acquired is 6 m/s.

★ Finding distance travelled (s) ::

⚘ Using third eqⁿ of motion;

• $\boxed{\underline{\underline{\bf{\pink{v^2 = u^2 + 2as}}}}}\:\bf{\dag}$

Putting all values,

$\\ \longrightarrow\:\tt (6)^2 = (0)^2 + 2(0.02)s$

$\\ \longrightarrow\:\tt 36 = 0 + (2\:\times\:0.02)s$

$\\ \longrightarrow\:\tt 36 = 0 + 0.04s$

$\\ \longrightarrow\:\tt 36 = 0.04s$

$\\ \longrightarrow\:\tt \dfrac{36}{0.04} = s$

$\\ \longrightarrow\:\tt s = \dfrac{36\:\times\:100}{4}$

$\\ \longrightarrow\:\tt s = {\cancel{\dfrac{3600}{4}}}$

$\\ \longrightarrow\:{\underline{\boxed{\tt{s = 900\:m}}}}\:\bigstar$

• Hence, distance travelled is 900 m.

Learn more:

✧ Three equations of motion ✧

1. v = u + at
2. s = ut + ½ at²
3. v² = u² + 2as

✧ Some important definitions ✧

• Acceleration

Acceleration is the process where velocity changes. Since, velocity is the speed and it has some direction. So, change in velocity is considered as acceleration.

• Initial velocity

Initial velocity is the velocity of the object before the effect of acceleration.

• Final velocity

After the effect of acceleration, velocity of the object changes. The new velocity gained by the object is known as final velocity.

• Distance travelled

The total path length covered by an object is called distance travelled.

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