remarks the answers are easy to check (step 6). for example, check the answer to (b) by substituting known…

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

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

Answer

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