The-acceleration-of-a-particle-moving-in-a-straight-line-is-defined-as-a-6t-20-m-s-2-where-t-is-in-seconds-Knowing-that-s-0m-when-t-3s-and-that-t-5sec-when-v-2m-s-Determine-the-total-distance-trav

FacebookTweetPin

Question Number 69827 by Learner-123 last updated on 28/Sep/19

$$Theaccelerationofaparticlemoving$$$$inastraightlineisdefinedasa=6t-20$$$$m/s^2,wheretisinseconds.Knowing$$$$thats=0mwhent=3sandthatt=5sec$$$$whenv=2m/s.Determinethetotal$$$$distancetravelledwhent=11s.$$

Answered by Rio Michael last updated on 28/Sep/19

$$v=\int adt$$$$=\int (6t-20)dt=3t^2-20t+c$$$$v=3t^2-20t+cbutt=5,v=2$$$$2=5-100+c$$$$\Rightarrow c=2-5+100$$$$c=97$$$$\Rightarrow v=3t^2-20t+97$$$$s=\int vdt$$$$=\int (3t^2-20t+97)dt$$$$=t^3-10t^2+97t+c$$$$s=0att=3$$$$solveandsubstitudefort=11s$$

Commented by Learner-123 last updated on 28/Sep/19

$$Butthenyouwillgetdisplacement$$$$notdistance.$$

Commented by Rio Michael last updated on 29/Sep/19

$$yeahyourright$$

Answered by mr W last updated on 29/Sep/19

$$a=\frac{dv}{dt}=6t-20$$$${\int }_2^{v}dv={\int }_5^t(6t-20)dt$$$$v-2=3t^2-3\times 5^2-20(t-5)$$$$\Rightarrow v=\frac{ds}{dt}=3t^2-20t+27$$$${\int }_0^{s}ds={\int }_3^t(3t^2-20t+27)dt$$$$s=t^3-3^3-10(t^2-3^2)+27(t-3)$$$$\Rightarrow s=t^3-10t^2+27t-18$$$$att=0:s=-18$$$$att=11:s={11}^3-10\times {11}^2+27\times 11-18=400$$$$\mathrm{\Delta }s=400-(-18)=418m$$$$distancetravelled=418m$$

Commented by Learner-123 last updated on 29/Sep/19

$$thankssir.$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin