Part 1
A motorist traveling at 17 m/s encounters a deer in the road 48 m ahead.
If the maximum acceleration the vehicle’s brakes are capable of is −5 m/s^2, 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.
Part 2
If his or her reaction time is 1.29353s, how fast will (s) he be traveling when(s)he reaches the deer? Answer in units of m/s
I got that part 1 is 1.123s but i can't figure out part 2. Help??????
To solve Part 1 of the problem, we can use the equation of motion:
S = ut + (1/2)at^2
Where:
S = displacement (48 m)
u = initial velocity (17 m/s)
a = acceleration (-5 m/s^2)
t = time
Rearranging the equation, we get:
t = sqrt((2S) / a)
Substituting the given values:
t = sqrt((2 * 48 m) / -5 m/s^2)
Calculating this expression, we find that the maximum reaction time is approximately 1.123 seconds.
Moving on to Part 2 of the problem, we know the reaction time (t) is 1.29353 seconds. To determine the final velocity (v) when reaching the deer, we can use the equation:
v = u + at
Where:
u = initial velocity (17 m/s)
a = acceleration (-5 m/s^2)
t = time
Substituting the values into the equation:
v = 17 m/s + (-5 m/s^2 * 1.29353 s)
Calculating this expression, we find that the motorist will be traveling at approximately 10.5328 m/s when reaching the deer.