A bullet train start from rest and moves with unifrom acc 0.12m/s*2 find its final velocity and distance coverd after 5min
Answers
Answer:
Here u go dude
Explanation:
Time = 5 minutes. = (0.12 ×5×60) [∵ 1 minute = 60 seconds.] = 36 m/s. ∴ The final velocity of the train is 36 m/s.
Answer:
\mathtt{Initial \:velocity, u=0\:m/s }Initialvelocity,u=0m/s
\mathtt{Final\:velocity, yv=?}Finalvelocity,yv=?
\mathtt{Acceleration, a=0.12\:m/s²}Acceleration,a=0.12m/s²
\mathtt{Time,t=5min=5\times60=300\:sec}Time,t=5min=5×60=300sec
\mathtt{\red{\underline{Using \:first\: equation \:of \:motion\: }}}
Usingfirstequationofmotion
\mathtt{\red{\underline{to\: obtain\: the \:final \:speed}}}
toobtainthefinalspeed
\mathtt{v=u+at}v=u+at
\mathtt{v=0+0.12\times300=36m/s}v=0+0.12×300=36m/s
\mathtt{\underline{\blue{And\:the\: distance \:travelled\: is:}}}
Andthedistancetravelledis:
\mathtt{s = ut+\frac{1}{2}a{t}^{2}}s=ut+
2
1
at
2
= \mathtt{0 \times 300 + \frac{1}{2} \times 0.12 \times 300 \times 300}0×300+
2
1
×0.12×300×300
= \mathtt{ 5400 m }5400m
= \huge{\mathtt{5.4 km }}5.4km