[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
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[tex]\\[/tex]
[tex]\small{\blue{\text{Hikari\:Squad}}}[/tex]
