Question

A train travelling at 20 m second accelerate at 0.5 metre per second square for 30 second how far will it travel it is it how far will it travel in this time

Answer 1

We'll first jot down the given points. Initial velocity is 20m/s i.e. u. Final velocity would be zero since the bus stopped. It is noted that the bus deccelartion of 3m/s, so the acceleration should be considered negative of 3m/s I.e. -3m/s. So using third law of motion i.e.

v^2 - u^2= 2as

(0)^2 - (20)^2= -2×3×s {Note here I used minus sign in acceleration}

-400= -6×s

s=400÷6

s=66.67 m

So bus would cover a distance of 66.67m before coming to rest.

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Information :-

$\textbf{\underline{Initial\;velocity(u)}}$

= 20 m/s

$\textbf{\underline{Acceleration(a)}}$

= 0.5 m/s²

$\textbf{\underline{Time(t)}}$

= 30s

$\textbf{\underline{To\;Find}}$

Distance = ?

${\boxed{\sf\:{Using\;2^{nd}\;Equation\;of\;Motion}}}$

$\tt{\rightarrow s=ut+\dfrac{1}{2}\times at^2}$

$\textbf{\underline{Substituting\;the\;values}}$

$\tt{\rightarrow s=20\times 30+\dfrac{1}{2}\times 0.5\times (30)^2}$

$\tt{\rightarrow s=20\times 30+\dfrac{1}{2}\times 0.5\times 30\times 30}$

s = 600 + 0.5 × 15 × 30

s = 600 + 225

s = 825 m

$\textbf{\underline{Therefore}}$

${\boxed{\sf\:{Train\;will\;Cover\;825m}}}$

$\boxed{\begin{minipage}{9 cm} Additional Information \\ \\ \ Final\;Velocity=Initial\; velocity+Acceleration\times Time \\ \\ Momentum = Mass\times Velocity \\ \\ Force=Mass\times Acceleration \\ \\ Force = \dfrac{Change\;in\;Momentum}{Time\;Interval} \\ \\ Impulse = Force\times Time \end{minipage}}$