Question

A bullet train start from rest and moves with unifrom acc 0.12m/s*2 find its final velocity and distance coverd after 5min​

Answer 1

$\mathtt{Initial \:velocity, u=0\:m/s }$

$\mathtt{Final\:velocity, yv=?}$

$\mathtt{Acceleration, a=0.12\:m/s²}$

$\mathtt{Time,t=5min=5\times60=300\:sec}$

$\mathtt{\red{\underline{Using \:first\: equation \:of \:motion\: }}}$

$\mathtt{\red{\underline{to\: obtain\: the \:final \:speed}}}$

$\mathtt{v=u+at}$

$\mathtt{v=0+0.12\times300=36m/s}$

$\mathtt{\underline{\blue{And\:the\: distance \:travelled\: is:}}}$

$\mathtt{s = ut+\frac{1}{2}a{t}^{2}}$

= $\mathtt{0 \times 300 + \frac{1}{2} \times 0.12 \times 300 \times 300}$

= $\mathtt{ 5400 m }$

= $\huge{\mathtt{5.4 km }}$

Answer 2

Given :

• Initial velocity, u = 0 m/s (as it starts from rest)
• Acceleration, a = 0.12 m/s²
• Time taken, t = 5 minutes = 300 seconds

To find :

• Final velocity, v
• Distance covered

According to the question,

➞ v = u + at

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• t = Time taken

➞ Substituting the values,

➞ v = 0 + 0.12 × 300

➞ v = 0 + 36

➞ v = 36

• So, the final velocity is 36 m/s.

Now,

➞ s = ut + ½ at²

Where,

• s = Distance
• u = Initial velocity
• a = Acceleration
• t = Time taken

➞ Substituting the values,

➞ s = 0 × 300 + ½ × 0.12 × 300 × 300

➞ s = 0 + 36 × 150

➞ s = 0 + 5400

➞ s = 5400

• So,the distance covered is 5400 metres.