A motorist traveling at 15 m/s encounters a
deer in the road 39 m ahead.
If the maximum acceleration the vehicle’s
brakes are capable of is −6 m/s2, what is the
maximum reaction time of the motorist that
will allow her or him to avoid hitting the deer?
Answer in units of s
If his or her reaction time is 1.59 s, how fast
will (s) he be traveling when (s)he reaches the
deer?
Answer in units of m/s
To solve the first part of the question, we will use the kinematic equation that relates distance, initial velocity, acceleration, and time:
\[s = ut + \frac{1}{2}at^2\]
In this equation,
s = distance (39 m),
u = initial velocity (15 m/s),
a = acceleration (-6 m/s^2),
t = time (unknown).
We need to find the maximum reaction time, which means that the vehicle must stop just before hitting the deer (the final distance will be 0). So the equation becomes:
\[0 = ut + \frac{1}{2}at^2\]
Substituting the given values, we have:
\[0 = (15 \, \text{m/s})t + \frac{1}{2}(-6 \, \text{m/s}^2)t^2\]
Rearranging and simplifying, we get a quadratic equation:
\[-3t^2 + 15t = 0\]
Factoring out the common factor, we get:
\[t(-3t + 15) = 0\]
From this equation, we can see that t = 0 or t = 5.
Since we are looking for the maximum reaction time, we discard the t = 0 solution. Therefore, the maximum reaction time of the motorist is 5 seconds.
For the second part of the question, if the reaction time of the motorist is 1.59 seconds, we need to calculate the final velocity when reaching the deer. We can use the equation:
\[v = u + at\]
Where,
v = final velocity (unknown),
u = initial velocity (15 m/s),
a = acceleration (-6 m/s^2),
t = time (1.59 s).
Substituting the given values, we have:
\[v = 15 \, \text{m/s} + (-6 \, \text{m/s}^2)(1.59 \, \text{s})\]
Calculating this, we get:
\[v = 15 \, \text{m/s} - 9.54 \, \text{m/s} = 5.46 \, \text{m/s}\]
Therefore, the motorist will be traveling at a speed of 5.46 m/s when reaching the deer.