Question

Velocity of a particle increases from 10m/s to 15m/s after travelling a distance of 5m its acceleration is

Answer 1

Gɪᴠᴇɴ :-

• Initial velocity (u) = 10 m/s

• Final velocity (v) = 15 m/s

• Distance (s) = 5 m

ᴛᴏ ғɪɴᴅ :-

• Acceleration (a)

sᴏʟᴜᴛɪᴏɴ :-

✞ On Using 3rd Equation of motion , we get

➠ v² - u² = 2as

➠ (15)² - (10)² = 2×a × 5

➠ 225 - 100 = 10a

➠ 125 = 10a

➠ a = 125/10

➠ a = 12.5 m/s²

Hence,

• Acceleration (a) = 12.5 m/s²

➥ ᴍᴏʀᴇ ᴇǫᴜᴀᴛɪᴏɴ :-

❶ v = u + at

❷ s = ut + 1/2at²

❸ 2as = v² - u²

Answer 2

Given

Velocity of a particle increases from 10m/s to 15m/s after travelling a distance of 5m

To find

Find the acceleration

Solution

• Initial velocity (u) = 10m/s
• Final velocity (v) = 15m/s
• Distance (s) = 5m
• acceleration (a) = ?

*According to third equation of motion*

→ v² - u² = 2as

→ (15)² - (10)² = 2*a*5

→ 225 - 100 = 10a

→ 125 = 10a

→ a = 125/10 = 12.5m/s²

Hence, acceleration of particle is

12.5m/s²