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Juliette is driving her car when she sees a cat run across the road. If she is able to stop the car over a distance of 0.025km in 2.5 s, what is her average acceleration?

A. +10 m/s2
B. +0.01 m/s2
C. -0.004 m/s2
D. -4 m/s2

Sagot :

Luanaid
[tex]velocity=\frac{distance}{time}=\frac{0,025km}{2,5s}=\frac{25m}{2,5s}=100\frac{m}{s}\\\\ acceleration=\frac{velocity}{time}=\frac{-10\frac{m}{s}}{2,5s}=-4\frac{m}{s^2}\\\\Solution\ is\ D.[/tex]
Onyebuchinnajiid

The average acceleration of Juliette is - 4 m/s².

The given parameters;

  • distance traveled by the car, d = 0.025 km = 25 m
  • time of motion, t = 2.5 s

The initial velocity of Juliette is calculated as follows;

[tex]u = \frac{d}{t} \\\\u = \frac{25}{2.5} \\\\u = 10 \ m/s[/tex]

The average acceleration of Juliette is calculated as follows;

[tex]a = \frac{\Delta v}{t} \\\\a = \frac{v - u}{t} \\\\when\ she \ stops\ , the \ final \ velocity\ , v = 0\\\\a = \frac{0-10}{2.5} \\\\a = - 4 \ m/s^2[/tex]

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