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A taxi running at 30 m/s slowed down uniformly until it stopped in 44 s when a passenger signalled to it. Find the acceleration and the stopping distance.

Sagot :

AGboiiidn

[tex] \large\underline \mathcal{{QUESTION:}}[/tex]

A taxi running at 30 m/s slowed down uniformly until it stopped in 44 s when a passenger signalled to it. Find the acceleration and the stopping distance.

[tex]\\[/tex]

[tex] \large\underline \mathcal{{SOLUTION:}}[/tex]

Given that:

  • [tex]\sf{V_i=30m/s\:,\:V_f=0m/s\:,\:t=44sec}[/tex]

We can use the formula:

  • [tex]\sf{V_f=V_i+at}[/tex]

» Finding for the acceleration:

[tex]\sf{V_f=V_i+at}[/tex]

[tex]\sf{O=30m/s+a(44s)}[/tex]

[tex]\sf{-30m/s=44sa}[/tex]

[tex]\sf{\frac{30m/s}{44s}=a}[/tex]

[tex]\sf{-0.682m/s^2=a}[/tex]

[tex]\\[/tex]

» Finding for the stopping distance:

[tex]\sf{D=V_i(t)+\frac{1}{2}(a)(t^2)}[/tex]

[tex]\sf{D=30m/s(44s)+\frac{1}{2}(-0.682m/s^2)((44s)^2)}[/tex]

[tex]\sf{D=1320m-0.341(1936)}[/tex]

[tex]\sf{D=1320m-658.24}[/tex]

[tex]\sf{D=661.76m}[/tex]

[tex]\\[/tex]

[tex] \large\underline \mathcal{{ANSWER:}}[/tex]

Acceleration = -0.682m/s²

Stopping Distance = 661.76meters

[tex]\\[/tex]

[tex]\\[/tex]

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