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)
Answers
Answered by
4
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
Answered by
0
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..
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