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Question Number 30961 by Tinkutara last updated on 01/Mar/18

Commented by ajfour last updated on 01/Mar/18

$$\left(4\right)ifitdidnotstartfromrest.$$

Commented by MJS last updated on 01/Mar/18

$$\mathrm{no}.\mathrm{as}\mathrm{long}\mathrm{as}\mathrm{the}\mathrm{acc}.\mathrm{is}\mathrm{above}$$$$\mathrm{zero},\mathrm{the}\mathrm{speed}\mathrm{will}\mathrm{rise},\mathrm{no}\mathrm{matter}$$$$\mathrm{at}\mathrm{which}\mathrm{speed}\mathrm{you}\mathrm{start}$$

Commented by ajfour last updated on 01/Mar/18

$$startsfromoriginORstartsfrom$$$$restfromorigin,maybenotthe$$$$same.$$

Commented by Tinkutara last updated on 01/Mar/18

But how does it matter anyway? After 10 s it would be maximum because area under the curve will be maximum whether it starts from rest or not.

Commented by MJS last updated on 01/Mar/18

$$\mathrm{the}\mathrm{graph}\mathrm{shows}\mathrm{the}\mathrm{acceleration},$$$$\mathrm{not}\mathrm{the}\mathrm{velocity},\mathrm{so}\mathrm{the}\mathrm{answer}$$$$\mathrm{is}\left(2\right).{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{right}$$

Commented by Tinkutara last updated on 01/Mar/18

No it is only 4⃣. Now got it. Thanks ajfour Sir.

Commented by MJS last updated on 01/Mar/18

$$v_{10}=v_0+\underset{0}{\stackrel{10}{\int }}a\left(t\right)dt$$

Commented by ajfour last updated on 01/Mar/18

$$whatifinitialvelocityis$$$$sufficientlynegative,thendue$$$$topositiveacceleration,itwill$$$$slowdown..$$

Commented by Tinkutara last updated on 01/Mar/18

Yes I also thought this way and it became clear.

Commented by MJS last updated on 01/Mar/18

$$\mathrm{velocity}\mathrm{is}\mathrm{always}\mathrm{considered}\geqq 0,\mathrm{if}$$$$\mathrm{you}\mathrm{drive}80\mathrm{mph}\mathrm{backwards},$$$$\mathrm{braking}\mathrm{still}\mathrm{slows}\mathrm{you}\mathrm{down}.$$$$\mathrm{positive}\mathrm{acceleration}\mathrm{always}$$$$\mathrm{speeds}\mathrm{you}\mathrm{up}.$$$$\mathrm{if}\mathrm{the}\mathrm{given}\mathrm{example}\mathrm{would}\mathrm{deal}$$$$\mathrm{with}\mathrm{n}-\mathrm{dimensional}\mathrm{vectors},$$$$\mathrm{it}\mathrm{should}\mathrm{be}\mathrm{made}\mathrm{clear}\mathrm{somewhere}.$$$$\mathrm{but}\mathrm{then},\mathrm{the}\mathrm{answer}\mathrm{would}\mathrm{also}$$$$\mathrm{depend}\mathrm{on}\mathrm{the}\mathrm{direction}\mathrm{of}$$$$\mathrm{acceleration}.$$$$\mathrm{I}\mathrm{would}\mathrm{not}\mathrm{put}\mathrm{more}\mathrm{into}\mathrm{it}$$$$\mathrm{than}{\mathrm{what}}^{\prime }\mathrm{s}\mathrm{obviously}\mathrm{there}...$$

Commented by MJS last updated on 01/Mar/18

$$...\mathrm{Sir}\mathrm{Aifour},{\mathrm{something}}^{\prime }\mathrm{s}\mathrm{mixed}$$$$\mathrm{up}\mathrm{here}:\mathrm{if}{v}_0<0,\mathrm{still}{v}_0+a\geqq v_0$$$$\mathrm{with}a\geqq 0.\mathrm{but}:\mid v_0\mid \leqq \mid v_0+a\mid$$$$\mathrm{your}\mathrm{argument}\mathrm{uses}v\mathrm{and}\mid v\mid$$$$\mathrm{simultaneously}\mathrm{it}\mathrm{seems}$$

Commented by ajfour last updated on 01/Mar/18

$$ifsay{v}_0=-150m/s$$$$anda=t(10-t)$$$$thenv=-150+{\int }_0^{10}t(10-t)dt$$$$=-150+(500-\frac{1000}{3})=\frac{50}{3}m/s$$$$\Rightarrow speedhasdecreased.$$

Commented by MJS last updated on 01/Mar/18

$$\mathrm{I}\mathrm{see}\mathrm{what}\mathrm{you}\mathrm{mean}$$$$\mathrm{but}\mathrm{I}\mathrm{believe}\mathrm{the}\mathrm{given}\mathrm{problem}$$$${\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{meant}\mathrm{this}\mathrm{way}.\mathrm{a}\mathrm{moving}$$$$\mathrm{particle}\mathrm{accelerating}\mathrm{gets}\mathrm{faster}.$$$$\mathrm{any}\mathrm{other}\mathrm{interpretation}\mathrm{is}$$$$\mathrm{possible}\mathrm{but}\mathrm{seems}\mathrm{off}\mathrm{topic}.$$$$$$$$$$

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