Question

The speed of a particle varies as: v=2t²+t-10. The particle comes to rest momentarily at t equal to​

Answer 1

12:00 noon

0.54166666666667

0.45833333333333

0.58333333333333

Answer :

B

Solution :

v=2t3−3t2

dvdt=a=6t(t−1)

dvdt=0

t=0, 1sec

d2vdt2=12t−6⇒(d2vdt2)t=0=−6

(d2vdt2)t=1=6

At t=0, The sigh of double derivative is negative hence velocity will be maximum at 12:00 noon. At t=1, the sign of double derivative is positive hence the speed will be minimum at 1:00

P.M.