Question

It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it began to skid.

x = Question Blank 1 of 6
Answer

Vi = Question Blank 2 of 6
Answer

Vf = Question Blank 3 of 6
Answer

a = Question Blank 4 of 6
Answer

t = Question Blank 5 of 6
Answer

Answer 1

To find the initial speed (Vi) of the Jaguar before it began to skid, we can use the kinematic equation:

\[
V_f^2 = V_i^2 + 2a x
\]

where:
- \(V_f\) is the final velocity (0 m/s, since the Jaguar comes to a stop),
- \(V_i\) is the initial velocity (what we're trying to find),
- \(a\) is the acceleration (-3.90 m/s²),
- \(x\) is the distance (290 m).

Since the Jaguar comes to a stop, we know that:

\[
V_f = 0 \, \text{m/s}
\]

Plugging in the values and rearranging for \(V_i\):

\[
0 = V_i^2 + 2(-3.90)(290)
\]

Calculating:

\[
0 = V_i^2 - 2(3.90)(290)
\]
\[
0 = V_i^2 - 2262
\]
\[
V_i^2 = 2262
\]

Now, take the square root to find \(V_i\):

\[
V_i = \sqrt{2262} \approx 47.6 \, \text{m/s}
\]

Now we can summarize the results:

1. **x (length of skid)**: \(290 \, \text{m}\)
2. **Vi (initial speed)**: \(47.6 \, \text{m/s}\)
3. **Vf (final speed)**: \(0 \, \text{m/s}\)
4. **a (acceleration)**: \(-3.90 \, \text{m/s}^2\)
5. **t (time)**: We can find this using the equation:

\[
V_f = V_i + at
\]

Rearranging for time \(t\):

\[
t = \frac{V_f - V_i}{a}
\]
\[
t = \frac{0 - 47.6}{-3.90} \approx 12.2 \, \text{s}
\]

### Final Answers:
- **x**: \(290 \, \text{m}\)
- **Vi**: \(47.6 \, \text{m/s}\)
- **Vf**: \(0 \, \text{m/s}\)
- **a**: \(-3.90 \, \text{m/s}^2\)
- **t**: \(12.2 \, \text{s}\)