# A car has a mass of 710 kg. It starts from rest and travels 40.0 m in 3.0s. The car is uniformly accelerated during the entire time. How big is the net force acting on the car?

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Dec 4, 2015

$6300 \text{N}$

#### Explanation:

We will employ the following equation of kinematics

$x = \frac{1}{2} a {t}^{2}$

which describes an object traveling in one dimension with a constant acceleration and an initial velocity of zero.

We know the displacement and the time elapsed so we may solve for the acceleration:

40.0 "m" = 1/2a(3.0"s")^2

Solving for $a$ yields:

$a = 8.89 {\text{m"/"s}}^{2}$

Now, knowing the acceleration of the car as well as the mass of the car we can apply Newton's second law, which states

${F}_{\text{net}} = m a$

All we need to do is plug in our values for $m$ and $a$:

F_("net") = 710"kg"*8.89 "m"/"s"^2 = 6311.9 "N"

Considering the fact that our final answer should have only two significant digits, we will round to the nearest hundred:

F_("net") = 6300 "N"

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