Question

remarks the answers are easy to check (step 6). for example, check the answer to (b) by substituting known values into $delta x=\frac{1}{2}(v_0 + v)t$. for (d), an alternate technique is to use $delta x = v_0t+\frac{1}{2}at^{2}$ to find $t$ and then use $v = v_0+at$ to find $v$. also, notice in (c) that the two average - velocity calculations give slightly different answers. the difference is due to rounding and is not a cause for concern. question how fast is the car moving (in m/s) after covering a distance of 77.0 m from the starting line? 27.7 m/s exercise suppose the driver in this example slams on the brakes when the car is at $delta x = 45.0$ m, stopping the car in 5.10 s with constant acceleration. (a) find the acceleration (in $m/s^{2}$). 1.71 $m/s^{2}$ the response you submitted has the wrong sign. $m/s^{2}$ (b) find the distance (in m) the car travels while braking. 45 $m$ your response differs from the correct answer by more than 10%. double check your calculations. m resources read it

Explanation:

Step1: Identify the known - values

The car stops, so the final velocity $v = 0$. The car travels a distance $\Delta x=45.0$ m in time $t = 5.10$ s.

Step2: Use the kinematic equation $\Delta x=v_0t+\frac{1}{2}at^{2}$ and $v = v_0+at$ (where $v = 0$ so $v_0=-at$)

Substitute $v_0=-at$ into $\Delta x=v_0t+\frac{1}{2}at^{2}$, we get $\Delta x=-at^{2}+\frac{1}{2}at^{2}=-\frac{1}{2}at^{2}$.

Step3: Solve for acceleration $a$

$a=-\frac{2\Delta x}{t^{2}}$. Plugging in $\Delta x = 45.0$ m and $t = 5.10$ s, we have $a=-\frac{2\times45.0}{5.10^{2}}=-\frac{90}{26.01}\approx - 3.46$ m/s².

Step4: For part (b), since the car is decelerating with constant acceleration

We can also use the average - velocity formula $\Delta x=\frac{v_0 + v}{2}t$. Since $v = 0$, and from $v = v_0+at$ with $v = 0$ we know $v_0=-at$. Also, using the fact that the average velocity $\bar{v}=\frac{v_0 + v}{2}=\frac{v_0}{2}$. And we know from $v = v_0+at$ (with $v = 0$) that $v_0=-at$. The distance $\Delta x=\frac{v_0 + v}{2}t$. Since $v = 0$, $\Delta x=\frac{v_0}{2}t$. Another way is to use the kinematic equation $\Delta x=v_0t+\frac{1}{2}at^{2}$. Since $v_0=-at$, $\Delta x=-at^{2}+\frac{1}{2}at^{2}=-\frac{1}{2}at^{2}$. The distance the car travels while braking is the same as the given $\Delta x = 45.0$ m (as the problem is set up in terms of the braking distance). But if we calculate from the kinematic equations, using $v = 0$, $a=-3.46$ m/s² and $t = 5.10$ s and $v_0=-at$, $v_0=-(-3.46)\times5.10 = 17.65$ m/s. Then $\Delta x=\frac{v_0 + v}{2}t=\frac{17.65+0}{2}\times5.10=\frac{17.65\times5.10}{2}=45.0$ m.

Answer:

(a) $-3.46$ m/s²
(b) $45.0$ m