Question

A train is travelling at speed of 90km/hr
brakes are applied so as to produce
a uniform acceleration of -0.50/sec square Find
how far the train will go before it is
brought to rest​

Answer 1

$\huge{\underline{\sf{Solution:-}}}$

★ Speed of train:-

• u = 90 km/h = 90 × 5/18 = 25 m/s

★ Acceleration:-

• a = -0.5 m/s²

★ Using formula:-

• v = u + at

Finally train will be rest. So, final velocity,

• v = 0
• 0 = 25 - 0.5t
• 25 = 0.5t → t = 50 sec

★ Using formula:-

• S = ut + 1/2at²

Where S is distance travelled before stop

• S = 25 × 50 - 1/2 × 0.5 × 50²
• S = 1250 - 1/2 × 0.5 × 2500
• S = 1250 - 625
• S = 625 m

Hence, distance travelled = 625m and time taken = 50 sec.

Answer 2

$\tt{\huge{\green{Solution _{F.B } \::- }}}$

QUESTION :

A train is travelling at speed of 90km/hr .

Brakes are applied so as to produce an uniform acceleration of -0.50/sec square.

Find how far the train will go before it is brought to rest.

SOLUTION :

From the first equation of motion,

$\tt{ \purple{ \leadsto{a \: = \dfrac{v - u}{t} }}}$

U = 0 m/s.

V = 90 km/hr = 25 m/s.

Solving we get t = 50 Seconds.

Now,

From second Equation of motion,

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

U = 0

A = -0.5 m/s^2

T = 50 sec.

Solving we get S = 625 metres.