The-distance-x-m-have-used-by-a-particle-in-time-t-sec-is-described-by-the-equation-x-10-12t-2-Find-the-average-speed-of-the-particle-between-the-interval-t-2-sec-and-t-5-sec-

FacebookTweetPin

Question Number 9636 by tawakalitu last updated on 22/Dec/16

$$\mathrm{The}\mathrm{distance}\mathrm{x}\mathrm{m}\mathrm{have}\mathrm{used}\mathrm{by}\mathrm{a}\mathrm{particle}\mathrm{in}$$$$\mathrm{time}\mathrm{t}\sec \mathrm{is}\mathrm{described}\mathrm{by}\mathrm{the}\mathrm{equation}$$$$\mathrm{x}=10+{12\mathrm{t}}^2$$$$\mathrm{Find}\mathrm{the}\mathrm{average}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{particle}\mathrm{between}$$$$\mathrm{the}\mathrm{interval}\mathrm{t}=2\sec \mathrm{and}\mathrm{t}=5\sec$$

Commented by ridwan balatif last updated on 22/Dec/16

$$\mathrm{x}=10+{12\mathrm{t}}^2$$$$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=24\mathrm{t}$$$${\mathrm{v}}_1(\mathrm{t}=2\mathrm{s})=48\mathrm{m}/\mathrm{s}$$$${\mathrm{v}}_2(\mathrm{t}=5\mathrm{s})=120\mathrm{m}/\mathrm{s}$$$$\stackrel{-}{\mathrm{v}}=\frac{120-48}{5-2}=\frac{72}{3}=24\mathrm{m}/\mathrm{s}$$

Commented by tawakalitu last updated on 22/Dec/16

$$\mathrm{thanks}\mathrm{sir}.\mathrm{but}\mathrm{the}\mathrm{answer}\mathrm{options}\mathrm{here}\mathrm{are}.$$$$\mathrm{a})62\mathrm{m}/\mathrm{s}\left(\mathrm{b}\right)72\mathrm{m}/\mathrm{s}\left(\mathrm{c}\right)84\mathrm{m}/\mathrm{s}$$

Commented by ridwan balatif last updated on 22/Dec/16

$$\mathrm{ohh}{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{so}\mathrm{sorry},\mathrm{i}\mathrm{wrong}\mathrm{use}\mathrm{formula}$$$$\stackrel{-}{\mathrm{v}}=\frac{{\mathrm{x}}_2-{\mathrm{x}}_1}{{\mathrm{t}}_2-{\mathrm{t}}_1}$$$${\mathrm{x}}_1=10+12{\left(2\right)}^2$$$$=10+48=58\mathrm{m}$$$${\mathrm{x}}_2=10+12{\left(5\right)}^2$$$$=310\mathrm{m}$$$$\stackrel{-}{\mathrm{v}}=\frac{310-58}{5-2}$$$$=\frac{252}{3}$$$$=84\mathrm{m}/\mathrm{s}$$$$\mathrm{if}\mathrm{we}\mathrm{use}\mathrm{this}\mathrm{formula}\frac{{\mathrm{v}}_2-{\mathrm{v}}_1}{{\mathrm{t}}_2-{\mathrm{t}}_1},\mathrm{its}\mathrm{means}\mathrm{we}\mathrm{find}$$$$\mathrm{average}\mathrm{accelaration}$$

Commented by tawakalitu last updated on 22/Dec/16

$$\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{your}\mathrm{effort}\mathrm{sir}.$$$$\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Answered by mrW last updated on 22/Dec/16

$$\mathrm{distance}\mathrm{in}\mathrm{the}\mathrm{interval}\mathrm{t}=2\sec \mathrm{to}\mathrm{t}=5\sec :$$$$\mathrm{\Delta }\mathrm{s}=[10+12\times 5^2]-[10+12\times 2^2]=12\times (5^2-2^2)=252\mathrm{m}$$$$\mathrm{average}\mathrm{speed}\overline{\mathrm{v}}=\frac{\mathrm{\Delta }\mathrm{s}}{\mathrm{\Delta }\mathrm{t}}=\frac{252}{3}=84\mathrm{m}/\mathrm{s}$$

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

FacebookTweetPin