Question

a train accelerated from 20km/h to 80km/h in 4 min. how much distance does it cover in this period. assume that the tracks are straight?

Answer 1

Assuming the railway track are straight,

A train accelerated from 20 km/hr to 80km/hr in 4 minutes.

Given,

Time = 4 minutes = 4/60 hours = 1/15 hours.

Intital Velocity(u) = 20 km/hr

Final Velocity (v) = 80km/hr

We first find Acceleration (a)

Use first equation of motion,

v = u + at

⇒ a=v-u/t

⇒a = 80-20/1/15

⇒a = 60 × 15

⇒a = 900km/hr ²

Now, We find distance (s)

Use second equation of motion

s=ut+at²/2

⇒s =20(1/15)+900(1/15)(1/15)/2

⇒s =4/3+2

⇒s =10/3km

Therefore, The distance it covers in this period is 10/3 km

Answer 2

Answer: The train covers a distance of nearly 3.33 km in 4 min.

Explanation:

Initial velocity (u) = 20 km/hr  

Final Velocity (v) = 80 km/hr

Time (t) = 4min = $\frac{4}{60} =\frac{1}{15}$ hr

Now, the distance covered in 4 min by the train can be obtained by Newton’s equations of motion:

v² - u² = 2as .....(1)

Here, s is the displacement and a is the acceleration.

v² - u² = 2$2(\frac{v-u}{t})s$

We know that acceleration can be determined as,

a = $(\frac{v-u}{t} )$

Also, using the algebraic equation given below:

a²-b² = (a-b)(a+b)

The above equation can be simplified as,

(v-u) × (v+u) = 2$(\frac{v-u}{t} )s$

(v-u) × (v+u) × $(\frac{1}{(v-u)} )=2s$

(v+u)t × 2s

∴s = $\frac{(v+u)t}{2}$

s = $\frac{(20+80)}{2\times15}$

s = $\frac{100}{30}$

s = $\frac{10}{3}$

s = 3.33

Thus, the train covers a distance of nearly 3.33 km in 4 min.

#SPJ2