Question

the position of a particle is given by x=t^3-6t^2+9t. find its position at which velocity becomes zero

Answer 1

Explanation:

St=5ti^+6t2j^-10k^⇒Velocity=V=dSdt=5i^-12tj^⇒Acceleration

Answer 2

Answer:

x=t

3

−4t

2

+5t+9att=0x=9

v=3t

2

−8t+5=0

t=

6

8±

64−60

=

6

8±2

=

6

10

,1=

3

5

,1

att=1

x=1−4+5+9=11

att=3

x=

27

125

−

9

4×25

+

3

25

+9

=10.85

∴Totaldist=(11−9)+(11−10.85)=2.15m

Hence,

option (B) is correct answer.