Question

A train accelates uniformly from 36km/hr to 72km/hr in 20s find distance

Answer 1

u=36km/h =10m/s
v=72km/h =20m/s
t=20 s

a=v-u/t
=20-10/20
=0.5 m/s

now,
s=ut+1/2at^2
=10*20+1/2*0.5*20*20
=200+1/2*5/10*400
=200+5*20
=200+100
=300m




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Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Distance=300\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a train accelates uniformly from 36km/hr to 72km/hr in 20s.

• We have to find Distance travelled.

$\green{\underline\bold{Given :}} \\ :\implies \text{Initial\:velocity(u)=36\:km/h} \\ \\ :\implies \text{Final \: velocity (v)= 72 \:km/h} \\ \\ :\implies \text{Time= 20\:sec}\\ \\ \red{\underline \bold{To \: Find : }} \\ :\implies \text{Distance(s)= ?}$

• According to given question :

$\text{Initial\:velocity}=36\times \frac{5}{18}=10\:m/s\\\\ \text{Final\:velocity}=72\times \frac{5}{18}=20\:m/s\\\\ \bold{Using \: first \: equation \: of \: motion : } \\ : \implies v = u + at \\ \\ : \implies 20= 10 + a \times 20 \\ \\ : \implies 20\times a = 20-10 \\ \\ : \implies a= \frac{10}{20} \\ \\ \green{ : \implies {\text{a= 0.5\:m}/s^{2}}}$

$\bold{using \: second \: equation \: of \: motion } \\ : \implies s = ut + \frac{1}{2}a {t}^{2} \\ \\ : \implies s = 10 \times 20 + \frac{1}{2} \times \frac{1}{2} \times {20}^{2} \\ \\ : \implies s = 200 + \frac{1}{4} \times 400 \\ \\ : \implies s = 200 + 100 \\ \\ \green{ : \implies \text{ s = 300 \: m}}$