Question

a train starting from the rest moving with a uniform acceleration of 0.2 metre per second square for 5 minutes calculate the final velocity acquired and the distance travelled in the time​

Answer 1

acceleration, a = 0.2 m/s²

Initial velocity, u = 0 m/s (rest)

time taken, t = 5 min. = 300 sec.

We know that,

v = u + at

v = 0 + 0.2(300)

v = 60 m/s ____________(Ans.)

And we also know that

Speed = distance/time

Distance = speed x time

Distance = 60 x 300

Distance = 18000 m

Hence, Distance = 18 km_______(Ans.)

Hope it helps.

^_°

Answer 2

$\bf{\underline{Given}}:-$

• Acceleration (a) = 0.2 m/s²

• Initial velocity (u) = 0 m/s

• Time (t) = 5 min = 5 × 60 = 300 sec

$\bf{\underline{To\:Find}}:-$

• Final velocity (v) = ?
• The distance travelled (s) = ?

$\bf{\underline{Solution}}:-$

Using first equation of motion,

v = u + at

⤇ v = 0 + 0.2 × 300

⤇ v = 0.2 × 300

⤇ v = 60 m/s

Hence,the final velocity acquired will be 60 m/s.

Now,

Using third equation of motion,

v² = u² + 2as

⤇ (60)² = (0)² + 2 × 0.2 × s

⤇ 3600 = 2 × 0.2 × s

⤇ 3600 = 0.4s

⤇ $\rm\dfrac{3600}{0.4}=s$

⤇ 9000 = s

⤇ s = 9000 m or 9 km

Hence,the distance travelled will be 9000m or 9 km.

More information,

⤇ Velocity :- It is defined as the displacement covered by it's per unit time.

• It is a vector quantity.
• SI unit of velocity is m/s.
• CGS unit of velocity is cm/s.
• It also describe motion.
• velocity = Displacement/Time.