Answer:
0.19m/s²
Explanation:
Initial velocity(u) = 50×1000/60×60
=13.88 m/s
Final velocity(v) = 36.5×1000/60×60
=10.13 m/s
Acceleration(a) = v-u/t
=10.13-13.88/19.5
a= -0.19m/s²
-a = 0.19m/s²
The magnitude of retar dation is 0.19m/s²
Answer:
[tex]\sf 0.192\:ms^{-2}\:\:(3\:s.f.)[/tex]
Explanation:
Given:
First, convert the velocities into meters per second by dividing by 3.6:
[tex]\implies \sf 50.0\: km h^{-1}=\dfrac{50.0}{3.6}\:ms^{-1}= \dfrac{125}{9}\:ms^{-1}[/tex]
[tex]\implies \sf 36.5\: km h^{-1}=\dfrac{36.5}{3.6}\:ms^{-1}= \dfrac{365}{36}\:ms^{-1}[/tex]
To find the vehicle's acceleration, use one of the constant acceleration equations with the given values:
[tex]\implies \sf v=u+at[/tex]
[tex]\implies \sf \dfrac{365}{36}=\dfrac{125}{9}+19.5a[/tex]
[tex]\implies \sf \dfrac{365}{36}-\dfrac{125}{9}=19.5a[/tex]
[tex]\implies \sf 19.5a=-3.75[/tex]
[tex]\implies \sf a=-\dfrac{3.75}{19.5}[/tex]
[tex]\implies \sf a=-0.1923076923...\:ms^{-2}[/tex]
Therefore, the magnitude of acceleration is:
[tex]\implies \sf |a|=|-0.19230...|=0.192\:ms^{-2}\:\:(3\:s.f.)[/tex]
Learn more about Constant Acceleration Equations here:
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