A motorist suddenly notices a stalled car and slams on the brakes, negatively accelerating at \(6.3 \mathrm{m} / \mathrm{s}^{2} .\) Unfortunately, this isn't enough, and a collision ensues. From the damage sustained, police estimate that the car was going \(18 \mathrm{km} / \mathrm{h}\) at the time of the collision. They also measure skid marks 34 m long. (a) How fast was the motorist going when the brakes were first applied? (b) How much time elapsed from the initial braking to the collision?

Short Answer

Expert verified
The initial velocity is calculated in Step 4 and the time elapsed from the initial breaking to the collision is calculated in Step 3. Both are dependent on solving the quadratic equation correctly in Step 3.

Step by step solution

01

Convert final velocity to m/s

The final velocity provided is in km/h. Let's convert it to m/s using the conversion factor 1 km/h = 0.2778 m/s. Therefore, \(18 km/h = 18 * 0.2778 = 5 m/s\)
02

Find initial velocity (u)

We know that \(v= u - at\), where a is actually -6.3 m/s^2 (negative as it is deceleration), and v is 5 m/s we calculated above. We can rearrange this equation to solve for u, getting: \(u = v + at\). But to use this equation, we need to know 't'. Let's find 't' first.
03

Find Time

Using the equation of motion \(s = ut + 0.5at^2\), we can rearrange the equation to solve for t by plugging in known values (s=34m, v=5m/s and a=-6.3m/s^2), yielding a quadratic equation: \(t^2 + (v/a)t - (s/a) = 0\). Solving this quadratic equation will give us value of t.
04

Insert t into the velocity equation

Now that we have t, we can insert the t value and known values of v and a, into the velocity equation \(u = v + at\) to find the initial velocity, u.
05

Conclusion

Now we have the initial velocity and the time elapsed from the initial breaking to the collision.

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Most popular questions from this chapter

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