To determine the net force acting on the car, we need to follow these steps:
1. Find the acceleration of the car:
Given the initial velocity [tex]\( u = 0 \,\text{m/s} \)[/tex] (since the car starts from rest), the final velocity [tex]\( v = 20 \,\text{m/s} \)[/tex], and the time [tex]\( t = 5 \,\text{s} \)[/tex], we can use the formula for acceleration:
[tex]\[
a = \frac{v - u}{t}
\][/tex]
Substituting the given values:
[tex]\[
a = \frac{20 \,\text{m/s} - 0 \,\text{m/s}}{5 \,\text{s}} = \frac{20 \,\text{m/s}}{5 \,\text{s}} = 4 \,\text{m/s}^2
\][/tex]
2. Calculate the net force:
According to Newton's second law, the net force [tex]\( F \)[/tex] acting on an object is the product of its mass [tex]\( m \)[/tex] and its acceleration [tex]\( a \)[/tex]:
[tex]\[
F = m \cdot a
\][/tex]
Given the mass of the car [tex]\( m = 1.2 \times 10^3 \,\text{kg} \)[/tex] and the acceleration [tex]\( a = 4 \,\text{m/s}^2 \)[/tex], we can substitute these values into the formula:
[tex]\[
F = 1.2 \times 10^3 \,\text{kg} \cdot 4 \,\text{m/s}^2 = 4.8 \times 10^3 \,\text{N}
\][/tex]
3. Match the calculated force to the given options:
The options are:
1. [tex]\( 3.0 \times 10^2 \,\text{N} \)[/tex]
2. [tex]\( 60 \times 10^2 \,\text{N} \)[/tex]
3. [tex]\( 12 \times 10^3 \,\text{N} \)[/tex]
4. [tex]\( 28 \times 10^3 \,\text{N} \)[/tex]
5. [tex]\( 1.2 \times 10^2 \,\text{N} \)[/tex]
Our calculated value of [tex]\( 4.8 \times 10^3 \,\text{N} \)[/tex] fits none of the given options accurately. However, none of the options can be exactly equivalent to the correct value [tex]\( 4.8 \times 10^3 \,\text{N} \)[/tex]. For an ambiguous set of answer choices that involve hidden assumptions, I might need confirmation but considering closest values here, none is close to [tex]\( 4.8 \times 10^3\)[/tex].
Therefore, the best approach would be to reevaluate or correct any missing assumption.
Hence, reverify the entire working above clearly indicates [tex]\(4.8x10^3\inconsistency\)[/tex].
For forced choices - the correct contextual numerics’ involves verified confirmation.
