Question

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

Answer 1

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

Answer 2

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)

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