A motorist avoiding hitting a deer

[hsbc2003]

Homework Statement

Part 1:
A motorist is traveling at 17 m/s when he sees a deer in the road 45 m ahead.
If the maximum negative acceleration of vehicle is -7 m/s^2, what is the maximum reaction time (deltat) of the motorist that will allow him to avoid hitting the deer? Answer in units of s.

Part 2:
If his reaction time is 1.71654 s, how fast will he be traveling when he reaches the deer? Answer in units of m/s.

The Attempt at a Solution

I tried to use this equation: X(t)= X (initial)+ V*t + 0.5*a*t^2
but it does not work and I stopped
I can not think of a method to solve it

Please help

Thanks

 

[tiny-tim]

Oh deer!

A motorist is traveling at 17 m/s …

hi hsbc2003!

Hint: Start by working out how far he travels between starting to brake and coming to rest.

Everything else should then become clear.

 

[baba89]

for reaching out for help with this problem. I understand the importance of finding a solution to a problem and I am happy to assist you.

For Part 1, we can use the kinematic equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the final velocity would be 0 m/s (since the motorist wants to come to a stop to avoid hitting the deer), the initial velocity is 17 m/s, and the acceleration is -7 m/s^2. We can rearrange the equation to solve for time: t = (v - u)/a. Plugging in the values, we get t = (0 - 17)/-7 = 2.43 s.

Therefore, the maximum reaction time of the motorist that will allow him to avoid hitting the deer is 2.43 seconds.

For Part 2, we can use the same kinematic equation, v = u + at, but this time we are solving for the final velocity. We know that the initial velocity is 17 m/s, the acceleration is still -7 m/s^2, and the time is given as 1.71654 s. Plugging in the values, we get v = 17 + (-7)(1.71654) = 4.82 m/s.

Therefore, when the motorist reaches the deer, he will be traveling at a speed of 4.82 m/s.

I hope this helps and feel free to reach out if you have any further questions or need clarification. Keep up the good work as a scientist!