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Question Number 27435 by NECx last updated on 06/Jan/18
An object has a constant  acceleration a=4ms^(−2) .Its velocity  is 1m/s at t=0,when it is at x=7m.  How fast is it at x=8m?At what  time is this?
$${An}\:{object}\:{has}\:{a}\:{constant} \\ $$$${acceleration}\:{a}=\mathrm{4}{ms}^{−\mathrm{2}} .{Its}\:{velocity} \\ $$$${is}\:\mathrm{1}{m}/{s}\:{at}\:{t}=\mathrm{0},{when}\:{it}\:{is}\:{at}\:{x}=\mathrm{7}{m}. \\ $$$${How}\:{fast}\:{is}\:{it}\:{at}\:{x}=\mathrm{8}{m}?{At}\:{what} \\ $$$${time}\:{is}\:{this}? \\ $$
Answered by mrW1 last updated on 07/Jan/18
v(t)=v_0 +at  x(t)=x_0 +v_0 t+(1/2)at  where x_0 =7m, v_0 =1m/s.  8=7+1×t+(1/2)×4×t^2   ⇒2t^2 +t−1=0  ⇒t=((−1+(√(1^2 +4×2×1)))/4)=0.5 s  ⇒v=1+4×0.5=3 m/s    ⇒at x=8m, its speed is 3m/s and  the time is 0.5 second.
$${v}\left({t}\right)={v}_{\mathrm{0}} +{at} \\ $$$${x}\left({t}\right)={x}_{\mathrm{0}} +{v}_{\mathrm{0}} {t}+\frac{\mathrm{1}}{\mathrm{2}}{at} \\ $$$${where}\:{x}_{\mathrm{0}} =\mathrm{7}{m},\:{v}_{\mathrm{0}} =\mathrm{1}{m}/{s}. \\ $$$$\mathrm{8}=\mathrm{7}+\mathrm{1}×{t}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}×{t}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{t}^{\mathrm{2}} +{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{−\mathrm{1}+\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{4}×\mathrm{2}×\mathrm{1}}}{\mathrm{4}}=\mathrm{0}.\mathrm{5}\:{s} \\ $$$$\Rightarrow{v}=\mathrm{1}+\mathrm{4}×\mathrm{0}.\mathrm{5}=\mathrm{3}\:{m}/{s} \\ $$$$ \\ $$$$\Rightarrow{at}\:{x}=\mathrm{8}{m},\:{its}\:{speed}\:{is}\:\mathrm{3}{m}/{s}\:{and} \\ $$$${the}\:{time}\:{is}\:\mathrm{0}.\mathrm{5}\:{second}. \\ $$
Commented by NECx last updated on 07/Jan/18
thanks boss.
$${thanks}\:{boss}. \\ $$

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