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san4es73 [151]
3 years ago
5

A truck traveling with an initial velocity of 22m/s comes to a stop in 17.32 secs. What is the acceleration of the truck?

Physics
2 answers:
Tomtit [17]3 years ago
6 0

Explanation:

Given:

v₀ = 22 m/s

v = 0 m/s

t = 17.32 s

Find: a

v = at + v₀

(0 m/s) = a (17.32 s) + (22 m/s)

a = -1.270 m/s²

Round as needed.

trasher [3.6K]3 years ago
6 0

<u>Answer:</u>

A truck traveling with an initial velocity of 22 m/s comes to a stop in 17.32 sec, the acceleration of the truck is - 1.27 m/s^2

<u>Explanation</u>

From the Given statements, we know that Initial Velocity (u) = 22 m/s

time (t) = 17.32 seconds

Final Velocity (v) = 0 m/s [ because the truck Stops]

find acceleration on Applying the Equation of Motion

v = u + at

substituting the values in above equation, we find

  a   =\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}

       =\frac{0-22}{17.32}  \mathrm{m} / \mathrm{s}^{2}

 a  = - 1.27 m/s^2 [negative sign shows reverse direction]

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Now we can substitute the provided values into our equation. So we get:

\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}

so:

\frac{dh}{dt}=0.25m/s

3 0
3 years ago
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