Question

A ball starting from the rest hits the wall with the velocity 40 m/s. Find the distance covered by the ball in 4 seconds​

Answer 1

Aɴsᴡᴇʀ:

• The distance covered by the ball in 4 seconds = 80 m

Gɪᴠᴇɴ:

• Initial velocity (u) = 0
• Final velocity (v) = 40m/s
• Time (t) = 4 seconds

Nᴇᴇᴅ Tᴏ Fɪɴᴅ:

• The distance covered by the ball in 4 seconds = ?

Solution:

Bʏ ᴛʜᴇ ғɪʀsᴛ ᴇǫᴜᴀᴛɪᴏɴ ᴏғ ᴍᴏᴛɪᴏɴ:

• v = u + at

Pᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇs:

➦ 40 = 0 + a × 4

➦ 40 = 4a

➦ a = 40/4

➦ a = 10 m/s²

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Now,

Bʏ ᴛʜᴇ sᴇᴄᴏɴᴅ ᴇǫᴜᴀᴛɪᴏɴ ᴏғ ᴍᴏᴛɪᴏɴ:

• s = ut + 1/2 at²

Pᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇs:

➦ s = 0 × 4 + 1/2 × 10 × (4)²

➦ s = 1/2 × 10 × 16

➦ s = 5 × 16

➦ s = 80 m

Tʜᴇʀᴇғᴏʀᴇ:

• The distance covered by ball = 80 m

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Answer 2

$\sf{Given} \begin{cases} \sf{ Initial\: Velocity (u) = 0 } \\ \sf{ Final \: Velocity (v) = 40 \: m/s } \\ \sf{ Time (t) = 4s} \end{cases}$

To Find:

• Find the distance covered by Ball?

Formula Used:

• $\\ {\boxed{\sf{ v = u + at }}}$

• $\\ {\boxed{\sf{ s = ut + \dfrac{1}{2} at^2 }}}$

Solution:

To find the Acceleration;

We have to apply 1st Motion of Equation;

$\bullet{\boxed{\sf\pink{ v = u + at }}}$

➩ v = u + at

➩ 40 = 0 + a × 4

➩ 40 = 4a

➩ a = 40/4

➩ a = 10 m/s²

Then,

We have to Find the distance travelled by the ball in 4s;

$\bullet {\boxed{\sf\pink{ s = ut + \dfrac{1}{2} at^2 }}}$

➩ s = ut + ½ × at²

➩ s = 0 × 4 + ½ × 10 × 4²

➩ s = 0 + ½ × 10 × 16

➩ s = 10 × 8

➩ s = 80 m

Hence,

The Distance travelled by Ball in 4 sec is 80 m.