Question

starting from rest a train attain speed 30m/s in 10 sec .find its acceleration and distance covered during this time.​

Answer 1

Answer:

3 m/s² = Acceleration

150 m = Distance

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 0 m/s (As it starts from rest)
• Final velocity (v) = 30 m/s
• Time taken (t) = 10 s

We are asked to calculate acceleration and distance covered during this time.

Finding acceleration :

Acceleration is the rate of change in velocity. It is a vector quantity, that means it requires both magnitude and direction for its description. Negative acceleration is called retardation. Its SI unit is m/s².

By using the first equation of motion,

⇒ v = u + at

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

⇒ 30 = 0 + 10a

⇒ 30 - 0 = 10a

⇒ 30 = 10a

⇒ $\cancel{\dfrac{30}{10}}$ = a

⇒ 3 m/s² = a

∴ Acceleration is 3 m/s² .

Finding distance travelled :

By using the third equation of motion,

⇒ v² - u² = 2as

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

⇒ (30)² - (0)² = 2 × 3 × s

⇒ 900 - 0 = 6s

⇒ 900 = 6s

⇒ $\cancel{\dfrac{900}{6}}$ = s

⇒ 150 m = s

∴ Distance travelled during this time is 150 m.