A train normally travels at a uniform speed of 70 km/h on a long stretch of stra

Question

A train normally travels at a uniform speed of 70 km/h on a long stretch of straight. level track. On a particular day, the train mutt make a 2.20 minute stop at a stallion along thin track. If the train decelerates at a uniform role of 1.7 m/s^2 und accelerates not a role of 0.4 in/s^2, how much time is lost in stopping at the station? A ball it thrown vertically downward from the edge of a bridge 26 0 m high with an initial speed of 12.0 m/s. The ball falls all the way down and strikes the water below. Determine the magnitude of the velocity of the ball just before it strikes the water. m/s A photographer in a helicopter ascending vertically at a constant rate of 1.2 m/s accidentally drops a camera out the window when the helicopter is 64.0 m above the ground, (a) How long will it take the camera to reach the ground?

Explanation / Answer

Given

Normal velocity                N = 70 km/h = 70 x 0.278 = 19.44 m/s

Deceleration                      a’ = 1.7m/s2

Acceleration                       a = 0.4 m/s2

Duration of rest                TR = 2.20 minutes = 132 s

Solution

When the train stops its velocity becomes 0 from N

Using the equation of motion

v = u – a’t’

0 = 19.44 – 1.7xt’

t’ = 19.44/1.7

t’ = 11.44 s

The distance moved while stopping

s’ = ut’- ½ a’t’2

s’ = 19.44x11.44 – (1.7x11.442)/2

s’ = 111.15 m

When the train acetates from the rest to normal speed N

v = u + at

19.44 = 0 + 0.4t

t = 19.44/0.4

t = 48.6 s

The distance moved while restarting

s = ut + ½ at2

S = 0x48.6 + (0.4x48.62)/2

S = 472.39 m

Total time taken for the stop = TR + t + t’

T = 132 + 48.6 + 11.44

T = 192.04 s

Total distance travelled

D = S + S’

D = 472.39 + 111.15

D = 583.54 m

The time it would take for the train to cross the distance D normally

TN = D/N

TN = 583.54/19.44

TN = 30.02 s

Time lost

TL = T – TN

TL = 192.04 - 30.02

Tl = 162.02 s

The time lost would be 162.02 seconds.

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