Question

If x = (3t3 + 2t2 + 2t + 7) m
and t = 5 sec.
find velocity, acceleration at t = 0, t = 1 sec​

Answer 1

Hii..

dy/dx=(9t^2+4t+20)m

1.displacement=9*0+0+20=20

velocity=20/5=4

2.displacement=9+4+20=33

velocity=33/5=6.6

accelration=6.6-4/5=2.6/5

Answer 2

V=ds/dt=9t^2+4t+2

velocity at o sec is 9(0)^2+4(0)+2=2m/s

velocity at 1sec is 9(1)^2+4(1)^2+2=15m/s

a=DV/dt=18t+4

acceleration at 0sec is 18(0)+4=4m/s^2

acceleration at 1sec is 18(1)+4=22m/s^2

hope you understand this.