Question

The displacement of a particle in time 't' is given by s=t³-t²-8t-18. The acceleration of the particle when it's velocity vanishes is (in units/sec²)

Answer 1

Answer:

10 units/sec²

Step-by-step explanation:

we know,

instantaneous velocity=ds/dt

=>d(t³-t²-8t-18)/dt

=>3t²-2t-8

when velocity vanishes,that is it becomes zero,

3t²-2t-8=0

3t(t-2)+4(t-2)=0

=>(t-2)(3t+4)=0

so,t=2 [t≠-4/3]

we know instantaneous acceleration=dv/dt

=d(3t²-2t-8)/dt

=6t-2

so,acceleration at t=2:

=6*2-2

=10 units/sec²

Answer 2

Answer:

hey mate your answer is

Step-by-step explanation:

a=v^3-u^2-8v-18u.-u^2m/s-8

( a) msec^2=v^3m/s-u^2m/s-8m/s-18m/s.

mark as brailiant answer..