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3 years ago

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the train accelerates from 30 km/h to 45 km/h in 15 secs. a. find its acceleration. b. distance it travels during this time ...

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2 answers:

Irina-Kira [14]3 years ago

4 0

Answer :

(a) The acceleration is, $0.278m/s^2$

(b) Distance it travels during this time is, 156.225 m

<u>Solution for part (a) :</u>

Formula used :

$a=\frac{v-u}{t}$

where,

a = acceleration

v = final velocity = $45Km/hr=45\times \frac{5}{18}=12.5m/s$

u = initial velocity = $30Km/hr=30\times \frac{5}{18}=8.33m/s$

t = time = 15 s

Now put all the given values in the above formula, we get

$a=\frac{(12.5-8.33)m/s}{15s}=0.278m/s^2$

Therefore, the acceleration is, $0.278m/s^2$

<u>Solution for part (b) :</u>

Formula used :

$s=u\times t+\frac{1}{2}a\times t^2$

where,

s = distance traveled

Now put all the given values in this formula, we get

$s=(8.33m/s)\times (15s)+\frac{1}{2}\times (0.278m/s^2)\times (15s)^2=156.225m$

Therefore, the distance it travels during this time is, 156.225 m

Sonbull [250]3 years ago

3 0

Acceleration = v-u/t
= (45-30)/15
= 1 km/h

Distance = ut + 1/2 at^2
s = [30 x 15] + 1/2 x 1 x 15^2
= 562.5 km

Hope this helps

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