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saul85 [17]
3 years ago
14

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 ...

?
Physics
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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