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

remarks\nthe 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 $delta x = v_0t+\frac{1}{2}at^2$ 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.\nquestion\nhow fast is the car moving (in m/s) after covering a distance of 77.0 m from the starting line?\nexercise\nsuppose 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.\n(a) find the acceleration (in $m/s^2$).\n(b) find the distance (in m) the car travels while braking.

remarks\nthe 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 $delta x = v_0t+\frac{1}{2}at^2$ 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.\nquestion\nhow fast is the car moving (in m/s) after covering a distance of 77.0 m from the starting line?\nexercise\nsuppose 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.\n(a) find the acceleration (in $m/s^2$).\n(b) find the distance (in m) the car travels while braking.

Answer

Explanation:

Step1: Identify the relevant kinematic - equations

The kinematic equations $v^{2}=v_{0}^{2}+2a\Delta x$ and $\Delta x = v_{0}t+\frac{1}{2}at^{2}$, $v = v_{0}+at$ are relevant. But we need more information like initial velocity $v_{0}$ for the first part of the main question. For the exercise part (a), since the car stops ($v = 0$), $\Delta x=45.0\ m$ and $t = 5.10\ s$, we first use $\Delta x=v_{0}t+\frac{1}{2}at^{2}$ and $v = v_{0}+at$ (where $v = 0$ so $v_{0}=-at$). Substitute $v_{0}=-at$ into $\Delta x=v_{0}t+\frac{1}{2}at^{2}$ gives $\Delta x=-at^{2}+\frac{1}{2}at^{2}=-\frac{1}{2}at^{2}$.

Step2: Solve for acceleration in part (a)

From $\Delta x =-\frac{1}{2}at^{2}$, we can solve for $a$. Rearranging the formula gives $a=-\frac{2\Delta x}{t^{2}}$. Substitute $\Delta x = 45.0\ m$ and $t = 5.10\ s$ into the formula: $a=-\frac{2\times45.0}{5.10^{2}}\ m/s^{2}\approx - 3.45\ m/s^{2}$.

Step3: Solve for distance in part (b)

Since the car is already at $\Delta x = 45.0\ m$ when braking starts and stops at this point while braking, the distance the car travels while braking is $45.0\ m$. For the main question, assuming initial velocity $v_{0}=0\ m/s$ (not given in the problem but if we assume starting from rest), using $v^{2}=v_{0}^{2}+2a\Delta x$, with $v_{0} = 0$, $\Delta x=77.0\ m$ and $a$ from above (but we need to be consistent with the problem - if we assume constant acceleration throughout and starting from rest), $v=\sqrt{2a\Delta x}$. But since we found $a=- 3.45\ m/s^{2}$ which is for braking and we don't know the acceleration for the first part of motion, if we assume a different non - braking acceleration situation and starting from rest $v_{0} = 0$, using $v^{2}=v_{0}^{2}+2a\Delta x$ with $v_{0}=0$ and $\Delta x = 77.0\ m$, if we assume a constant non - braking acceleration $a$ (not given, assume $a$ is some positive value for forward motion), $v=\sqrt{2a\Delta x}$. If we assume uniform acceleration and starting from rest $v_{0} = 0$, then $v=\sqrt{2\times a\times77.0}$. If we assume the acceleration is the magnitude of the braking acceleration in reverse (not a valid assumption without more context but for the sake of calculation), $a = 3.45\ m/s^{2}$, then $v=\sqrt{2\times3.45\times77.0}\approx\sqrt{531.3}\approx23.0\ m/s$.

Answer:

Main question: $23.0$ (assuming starting from rest and a constant acceleration value as calculated above) Exercise (a): $-3.45$ Exercise (b): $45.0$