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Exercise 6.2 · Q12

Q.A 150150 m long train is moving with constant velocity of 12.512.5 m/s. Find

(i) the equation of the motion of the train
(ii) the time taken to cross a pole
(iii) the time taken to cross a bridge of length 850850 m.
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Track the position of the REAR of the train (since "crossing" is complete only once the whole train has passed), starting the clock when the FRONT reaches the reference point; this gives a single linear equation that answers all three parts.

Let x=x= time in seconds measured from the instant the front of the train reaches a reference point (a pole, or the start of a bridge), and let y=y= the position (in metres) of the rear of the train relative to that same reference point. At x=0x=0 the front is at the reference point, so the rear — 150150 m behind the front — is at y=−150y=-150.

Step 1. Part (i) — equation of motion. The train moves at constant velocity 12.512.5 m/s, so yy increases at rate 12.512.5 per second, starting from y=−150y=-150 at x=0x=0 (point-slope through (0,−150)(0,-150) with slope 12.512.5):

y−(−150)=12.5(x−0)  ⇒  y=12.5x−150y-(-150)=12.5(x-0) \;\Rightarrow\; y=12.5x-150

Step 2. Part (ii) — time to cross a pole. A pole has (effectively) zero length, so crossing is complete the instant the rear reaches the pole, i.e. y=0y=0:

0=12.5x−150  ⇒  x=15012.5=12 seconds0=12.5x-150 \;\Rightarrow\; x=\frac{150}{12.5}=12\text{ seconds} …

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