Question

A train starting from rest attains a velocity of 72 km / h in 5 minutes. Assuming that the acceleration is uniform. find (i) the acceleration and (i) the distance traveled by the train for attaining this velocity -​

Answer 1

Given :-

Initial velocity (u) :- 0 m/s

Final velocity (v) :- 72 km/hr = 72 × 5/18 = 20 m/s

Time (t) :- 5 minutes = 5 × 60 seconds

To Find :-

Acceleration (a)

Distance travelled (s)

Solution :-

We know :-

$\bf v = u + at$

$: \implies \sf a = \dfrac{v - u}{t}$

$: \implies \sf a = \dfrac{20 - 0}{5 \times 60}$

$: \implies \sf a = \dfrac{20}{300}$

$: \implies \sf a = \dfrac{1}{15}$

Acceleration :- 1/15 m/s^2

We know :-

$\bf s = ut + \dfrac{1}{2} a {t}^{2}$

$: \implies \sf s = 0 \times 5 \times 30 + \dfrac{1}{2} \times \dfrac{1}{15} \times 5 \times 60 \times 5 \times 60$

$: \implies \sf s = 5 \times 2 \times 5 \times 60$

$: \implies \sf s = 3000$

Distance travelled :- 3000m = 3 km

Answer 2

Given :-

A train starting from rest attains a velocity of 72 km / h in 5 minutes. Assuming that the acceleration is uniform.

To Find :-

(i) Acceleration

(ii) distance traveled by the train for attaining this velocity -​

Solution :-

We know that

1 km/h = 5/18 m/s

72 km/h = 72 × 5/18

72 km/h = 4 × 5

72 km/h = 20 m/s

1 min = 60 sec

5 min = 5 × 60 = 300 sec

Now,

◼ F i n d i n g  A c c e l e r a t i o n :-

v = u + at

Where

v = final velocity

u = initial velocity

a = acceleration

t = time

⮆  20 = 0 + a(300)

⮆  20 = 0 + 300a

⮆  20 - 0 = 300a

⮆  20 = 300a

⮆ 20/300 = a

⮆ 2/30 = a

⮆ 1/15 = a

Acceleartion is 1/15 m/s²

◼ F i n d i n g  D i s t a n c e

v² - u² = 2as

Where

v = final velocity

u = initial velocity

a = acceleration

s = distance

⮆ (20)² - (0)² = 2(1/15)s

⮆ 400 = 2/15s

⮆ 400 × 15/2 = s

⮆ 200 × 15 = s

⮆ 3000 = s