Physics, asked by pritha16, 11 months ago

The distance travelled by a particle in time t is given by s = (2.5 t²) m. Find (a) the average speed of the particleduring time 0 to 5.0s and (b) the instantaneous speed at t = 5.0 s.
Please help me to find the answer and don't go for the points.

Answers

Answered by nitishmadhepura45
11

Answer:

a, <v>=12.5 m/s

b, ins. speed= 25m/s

Distance S=(2.5t^2)m.........(given)

Average speed of particals during time

0 to 5 sec.=Total distance /Total time

total distance=(2.5t^2)= 2.5× 5×5=62.5

total time=5 sec

<v>= 62.5/5= 12.5m/s

Therefore the average speed of the particle during time 0 to 5.0s is 12.5m/s.

Instantaneous speed=v = ds/dt =2.5t^2/dt =5t.

( Since at t=5) =25m/sTherefore the instantaneous speed at t=5.0s is 25m/s.

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