Question

1. a car is traveling down the 402 when it pulls out to pass a truck. the car accelerates for 4.0s reaching a velocity of 28m to pass the truck. if it takes 96m to pass the truck, what was the rate of acceleration?

Explanation:

Step1: Identify the relevant kinematic equation

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

Step2: Substitute $v_0$ into the displacement equation

Substitute $v_0 = v - at$ into $x = v_0t+\frac{1}{2}at^{2}$. We get $x=(v - at)t+\frac{1}{2}at^{2}$. Expand the right - hand side: $x = vt - at^{2}+\frac{1}{2}at^{2}=vt-\frac{1}{2}at^{2}$.

Step3: Rearrange the equation to solve for $a$

We are given that $x = 96m$, $v = 28m/s$, and $t = 4.0s$. Rearranging $x = vt-\frac{1}{2}at^{2}$ for $a$ gives $\frac{1}{2}at^{2}=vt - x$. Then $a=\frac{2(vt - x)}{t^{2}}$.

Step4: Plug in the values

Substitute $x = 96m$, $v = 28m/s$, and $t = 4.0s$ into the formula $a=\frac{2(vt - x)}{t^{2}}$. First, calculate $vt=28\times4 = 112m$. Then $vt - x=112 - 96 = 16m$. And $t^{2}=4^{2}=16s^{2}$. So $a=\frac{2\times16}{16}=2m/s^{2}$.

Answer:

$2m/s^{2}$