Question

please solve this qn.. ​

Answer 1

✪ᴀɴsᴡᴇʀ✪

• The train stops after 41.67 sec.

• Velocity after 20s of applying brake is 8.67 m/s.

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

☆ɢɪᴠᴇɴ☆

• Initial Velocity, u = 60 Km/Hr

• Acceleration, a = - 40cm/s^2

• Final Velocity, v = 0

☆ᴛᴏ ғɪɴᴅ☆

• Time when the train stops

• Velocity after 20s of applying brake.

☆ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ☆

Unit conversion

$\to u = 60 Km/Hr$

$\to u = 60\times \frac{1000\:m}{3600\:s}$

$\to u = \frac{50}{3}m/s$

And

$\to a = - 40cm/s^2$

$\to a = - 40\times {10}^{-2} m/s^2$

$\to a = - 0.4m/s^2$

⍟ᴛɪᴍᴇ ᴛᴀᴋᴇɴ ᴛᴏ sᴛᴏᴘ⍟

Using the given data in equation of motion

$\:\:\:\:\:\:\:\: v = u +at$

we get,

$\to 0 = \frac{50}{3} - (0.4)t$

$\to t = \frac{50}{3\times 0.4}$

$\to t = \frac{125}{3}$

$\to t = 41.67\:sec$

✫ᴠᴇʟᴏᴄɪᴛʏ✫

Velocity at t = 20 sec is given by

$\:\:\:\:\:\:\:\: v = u +at$

$\to v = \frac{50}{3} - (0.4)\times 20$

$\to v = \frac{50}{3} - 8$

$\to v = \frac{50-24}{3}$

$\to v = \frac{26}{3}$

$\to v = 8.67\:m/s$