train starting from rest moves with a uniform acceleration of 0-2 m/s* for 5 minutes. Calcualte the specd acquired and the distance travelled in this time.
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Answer:
train starting from rest moves with a uniform acceleration of 0.2m/s2 for 5 minutes . Calculate the speed acquired and the distance travelled in this time. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. Here, u=0,a=0.2m/s2,t=5min=5×60s=300s,v=?,s=?
explanation ::Initial velocity, u=0m/s
Initial velocity, u=0m/s Final velocity, yv=?
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300sec
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+at
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/s
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is s=ut+
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is s=ut+ 2
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is s=ut+ 21
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is s=ut+ 21
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is s=ut+ 21 αt
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is s=ut+ 21 αt 2
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is s=ut+ 21 αt 2
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is s=ut+ 21 αt 2 s=0×300+
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is s=ut+ 21 αt 2 s=0×300+ 2
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is s=ut+ 21 αt 2 s=0×300+ 21
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is s=ut+ 21 αt 2 s=0×300+ 21
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is s=ut+ 21 αt 2 s=0×300+ 21 ×0.2×300×300
Initial velocity, u=0m/s Final velocity, yv=?Acceleration, a=0.2m/s 2 Time, t=5min=5×60=300secUsing first equation of motion to obtain the final speed:v=u+atv=0+0.2×300=60m/sAnd the distance travelled is s=ut+ 21 αt 2 s=0×300+ 21 ×0.2×300×300s=0+9000=9000m=9km
Answer:
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