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Pani-rosa [81]
3 years ago
6

A car traveling at 32 m/s starts to decelerate steadily. It comes to a complete stop in 14 seconds. What is its acceleration?

Mathematics
2 answers:
Andrew [12]3 years ago
5 0

Hello!

A car traveling at 32 m/s starts to decelerate steadily. It comes to a complete stop in 14 seconds. What is its acceleration?

We have the following data:

Vi (initial velocity) = 32 m/s (starts)

Vf (final velocity) = 0 m/s (stop)

t (time) = 14 s

a (acceleration) = ? (in m/s²)

We apply the data to the formula of the hourly function of the velocity, let us see:

V_f = V_i + a*t

0 = 32 + a*14

- 32 = 14\:a

14\:a = - 32

a = \dfrac{-32}{14}

\boxed{\boxed{a = - 2.28\:m/s^2}}\Longrightarrow(the\:car\:slows\:down)\:\:\:\:\:\:\bf\green{\checkmark}

Answer:

The acceleration is -2.28 m/s² (decelerate)

________________________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}

valentina_108 [34]3 years ago
3 0
We have:

Initial velocity (u) = 32 m/s
Final velocity (v) = 0 m/s ⇒ The value is zero because the car comes to stationary position when it stops
Time = 14 seconds

We can use one of the constant acceleration equation:
v=u+at where a is the acceleration

0=32+14a
32=-14a
a=- \frac{32}{14}=-2.3m/s^{-2}

The acceleration is 2.3 m/s⁻² and the negative sign shows deceleration
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