Question

A train is running at a speed of 44 meters / second. After breaking, he stops at 121 meters. Find out the value of the retarded by the brake. How long it took to stop the train​

Answer 1

Initial Velocity of Train = 44 m/sec

Distance covered during retardation is 121 metres.

Since, the train is brought to rest:-

The value of final velocity will be 0 m/sec.

Now, we have :-

V=44m/sec

U=0m/sec

S=121 m

Applying the third equation of motion to find the retardation :-

$2as ={v}^{2} - {u}^{2} \\ \\ \\2 \times 121 \times a = - (44)(44) \\ \\ \\ \\2 \times 121a = - 44 \times 44 \\ \\ \\a = \frac{ - 44 \times 44}{2 \times 121} \\ \\ \\a = \frac{ - 4 \times 4}{2} \\ \\ \\a = - 8m \: per \: {sec}^{2}$

Now,

The time taken to stop the train can be calculated using the first equation of the motion :-

$v = u + at \\ \\ \\0 = 44 + ( - 8)t \\ \\ \\8t = 44 \\ \\ \\t = \frac{44}{8} \\ \\ \\t = 5.5 \: sec$

Thus time taken to stop the train was 5.5 seconds.

The retardation value was (-8)m/sec²

Answer 2

Intial Velocity (u) before Break are Applied=44 m/s

Final Velocity (v) after Train Stop=0 m/s

Train Travels Distance After Break are applied(s)=121 m

Use Equation of Motion

v²=u²+2as

0²=44²+2×a×121

0=1936+242a

-1936=242a

-1936/242=a

a=-8

Retardation or Negative Acceleration=-8 m/s²

It means that Velocity Of Train Decreasing at rate of 8m/s

Now We got

Intial Velocity (u)=44 m/s

Final Velocity (v)=0 m/s.

Acceleration (a)=-8 m/s²

Time(t)=?

Use Equation of Motion

v=u+at

0=44+-8×t

-44=-8t

t=-44/-8

t=5.5

$\boxed{\mathbf{\large{Retardation=-8\:m/{s}^{2}}}}$

$\boxed{\mathbf{\large{Time=5.5\:seconds}}}$