Question

A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of -0.5m/s^2. Find how far the train will the go before it is brought to rest.​

Answer 1

GIVEN

A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of -0.5m/s^2.

TO FIND

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

SOLUTION

• Initial velocity (u) = 90km/h
• Final velocity (v) = 0
• Acceleration (a) = -0.5m/s² {(-) minus shows retardation}
• Distance (s) = ?

→ u = 90km/h

→ u = 90 × 5/18

→ u = 25m/s

**According to the third equation of motion**

→ v² - u² = 2as

→ (0)² - (25)² = 2*(-0.5)*s

→ 0 - 625 = -s

→ - 625 = -s

→ s = 625m

Hence, 625m distance travelled by train before it is bought to rest

Extra Information

• v = u + at {First eqⁿ of motion}
• s = ut + ½ at²{Second eqⁿ of motion}

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Answer 2

Gɪᴠᴇɴ :-

• Initial velocity (u) = 90 km/ h × 5/18 = 25 m/s

• Final velocity (v) = 0. ( Train to be at rest)

• Acceleration (a) = -0.5 m/s² (Deceleration)

ᴛᴏ ғɪɴᴅ :-

• Distance travelled (s)

sᴏʟᴜᴛɪᴏɴ :-

By using Third equation of motion,

➡ v² - u² = 2as

where,

v = final velocity

u = initial velocity

a = acceleration

s = distance travelled

We get,

→ 0² - (25)² = 2× -0.5 × s

→ -625 = -1s

→ s = 625 m

Hence,

• Distance travelled (s) = 625 m