Question

5.  A train accelerated from 10km/hr to 40km/hr in 2 minutes. How much distance does it cover in this period? Assume that the tracks are straight?

Answer 1

Solution :

⏭ Given:

✏ Initial velocity of train = 10kmph

✏ Final velocity of train = 40kmph

✏ Time interval = 2min

⏭ To Find:

✏ Distance covered by train in that time interval

⏭ Formula:

✏ Formula of acceleration in terms of change in velocity and time interval us given by

$\bigstar \: \red{ \tt{a = \dfrac{v - u}{t}}}$

✏ Formula of distance covered in uniform acceleratory motion is given by

$\bigstar \: \tt{ \blue{s= ut + \dfrac{1}{2} a {t}^{2} }}$

⏭ Conversation:

✏ We know that 1kmph = 5/18mps

✒ 10kmph = 10 × 5/18 = 2.78mps

✒ 40kmph = 40 × 5/18 = 11.11mps

✒ 2min = 2 × 60 = 120s

⏭ Calculation:

$\implies \sf \: s = ut + \dfrac{1}{2} a {t}^{2} \\ \\ \implies \sf \: s = ut + \dfrac{1}{2} ( \dfrac{v - u}{t} ) {t}^{2} \\ \\ \implies \sf \: s = ut + \dfrac{1}{2} (v - u)t \\ \\ \implies \sf \: s = (2.78 \times 120) + \dfrac{1}{2} (11.11 - 2.78)(120) \\ \\ \implies \sf \: s = 333.6 + 499.8 \\ \\ \implies \: \boxed{ \tt{s = 833.4 \: m}}$

Answer 2

$\tt{\orange{\huge{Hello!!! }}} </p><p></p><p><strong>$Question :

5.  A train accelerated from 10km/hr to 40km/hr in 2 minutes. How much distance does it cover in this period? Assume that the tracks are straight?

Solution :

$a = \dfrac{v - u}{t} = \dfrac{8.3}{120} m. {s}^{ - 2} = 0.069 \: m. {s}^{ - 2}$

$s \: = ut \: + \dfrac{1}{2} a {t}^{2} = 833.4 \: m.$