Question

A train starting from rest attains a velocity of 36 km/h with 5minutes. Caluculate the distance to attain the velocity

Answer 1

Given:-

Initial Velocity (u) = 0m/s

Final Velocity (v) = 36km/h = 10m/s

Time Taken = 5min = 300s.

To find:-

The Distance train travelled to attain the final Velocity.

Formulae used:-

• v = u + at

• v² - u² = 2as.

Where,

v = Final Velocity

u = Initial Velocity

a = Acceleration

s = Distance

t = Time.

Now, Using first equation of motion to find accceleration.

v = u + at

or,

a = v - u/t

a = 10 - 0 / 300

a = 10/300

a = 0.03m/s²

Hence, The Acceleration is 0.03m/s².

Therefore, using Third equation of motion to find Distance travelled.

v² - u² = 2as

(10)² - (0)² = 2 × 0.03 × s

100 = 0.06 × s

S = 100/0.06

S = 1.666km.

Hence, The distance travelled by train is 1.6km to attain the Velocity.

Answer 2

Given

Initial velocity , u = 0 m/s

: Starts from rest :

Final velocity , v = 36 km/h = 10 m/s

Time , t = 5 min = 300 s

Answer

Acceleration is defined as rate of change in velocity

$:\to \bf a=\left(\dfrac{v-u}{t}\right)\ \; \pink{\bigstar}\\\\:\to \rm a=\dfrac{10-0}{300}\\\\:\to \rm a=\dfrac{10}{300}\\\\:\to \rm a=\dfrac{1}{30}\ m/s^2$

Apply 3rd equation of motion ,

$:\to \bf v^2-u^2=2as\ \; \orange{\bigstar}\\\\:\to \rm (10)^2-(0)^2=2\left( \dfrac{1}{30}\right)s\\\\:\to \rm 100-0=\dfrac{s}{15}\\\\:\to \rm 100=\dfrac{s}{15}\\\\:\to \rm s=1500\ m$

So , Distance = 1500 m