Math, asked by saqibi123890, 8 months ago

A train starts from rest and is accelerated with an acceleration . At time it decelerates with deceleration and finally stops after travelling total distance of 300 m.Find the distance travelled by train in time

Answers

Answered by HeroicGRANDmaster
2

Step-by-step explanation:

Total distance covered(s) = Distance during acceleration(s1) +  distance during uniform motion(s2) + distance during retardation(s3)

For acceleration,

v=u+a∗t

vmax=2×10=20 m/s

s=ut+(1/2)at2

s1=(1/2)×2×102

     =100 m

For uniform motion,

s=vt

s2=20×200

     =4000 m

During retardation,

v=u+at

0=20+50t

   =−0.4 m/s2

s3=ut+(1/2)at2

     =20×50+(1/2)×(−0.4)×502

     =500 m

Total distance travelled, s=s1+s2+s3=100+4000+500=4600 m

Total time taken, t=t1+t2+t3=260 s

Average velocity, vavg=Total TimeTotal Distance

                                      =4600/260=17.69 m/s

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