Question

• If the velocity of a train starting from rest becomes 72km/h in 10 minutes.

a) what is it's acceleration?
b) calculate the distance travelled by the train within this time interval..

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Answer 1

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A) What is it's acceleration?

==> Here , u = 0

==> v. = 72 km/h

$\frac{72 \times 1000}{60 \times 60} = 20 \:$

20 m/s

T = 10 minutes= 600 s

Acceleration a ==

$\frac{v - u}{t} = \frac{20ms - 0}{600 } = \frac{1}{30} m s{}^{2}$

So, Acceleration ==>

$\frac{1}{30} ms {}^{2}$

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B) Calculate the distance travelled by the train within this time interval.?

===>

${v}^{2} = {u}^{2} + 2as$

$(20) {}^{2} = 0 {}^{2} + 2 \times \frac{1}{30} \times s$

$400 = \frac{1}{15} \times s$

$s = 400 \times 15 = 6000m = 6 \: km$

Distance Travelled = 6 km