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A parked police car takes off after a suspect and reaches 81.0 km/h East in 8.05 seconds. What is the average acceleration (m/s2) of the car?

Sagot :

Given

v = 81 km/h

t = 8.05 s

Procedure

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).

[tex]\begin{gathered} a=\frac{\Delta v}{\Delta t} \\ \end{gathered}[/tex]

But first we need to convert the km/h of the velocity into m/s

[tex]\begin{gathered} 81\cdot\frac{\operatorname{km}}{\text{ h}}\cdot\frac{1000\text{ m}}{1\operatorname{km}}\cdot\frac{1\text{ h}}{3600\text{ s}} \\ 22.5\text{ m/s} \end{gathered}[/tex][tex]\begin{gathered} a=\frac{22.5m/s}{8.05s} \\ a=2.8m/s^2 \end{gathered}[/tex]

The acceleration would be 2.8 m/s^2