question 1 (3 points)\n(02.02 mc)\na car traveling 45.0 m/s starts to accelerate at 4.00 m/s². how far does…

question 1 (3 points)\n(02.02 mc)\na car traveling 45.0 m/s starts to accelerate at 4.00 m/s². how far does this car travel after 3.00 seconds?\n○ a 141 m\n○ b 147 m\n○ c 153 m\n○ d 204 m

question 1 (3 points)\n(02.02 mc)\na car traveling 45.0 m/s starts to accelerate at 4.00 m/s². how far does this car travel after 3.00 seconds?\n○ a 141 m\n○ b 147 m\n○ c 153 m\n○ d 204 m

Answer

Explanation:

Step1: Identify the kinematic - equation

We use the equation $x = v_0t+\frac{1}{2}at^{2}$, where $v_0$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.

Step2: Substitute the given values

Given $v_0 = 45.0\ m/s$, $a = 4.00\ m/s^{2}$, and $t = 3.00\ s$. $x=(45.0\ m/s)\times(3.00\ s)+\frac{1}{2}(4.00\ m/s^{2})\times(3.00\ s)^{2}$

Step3: Calculate each term

First term: $(45.0\ m/s)\times(3.00\ s)=135\ m$. Second term: $\frac{1}{2}(4.00\ m/s^{2})\times(3.00\ s)^{2}=\frac{1}{2}\times4.00\times9.00\ m = 18\ m$.

Step4: Sum the terms

$x=135\ m + 18\ m=153\ m$.

Answer:

C. 153 m