An engineer in a locomotive sees a car stuck

on the track at a railroad crossing in front of
the train. When the engineer first sees the
car, the locomotive is 170 m from the crossing
and its speed is 25 m/s.
If the engineer’s reaction time is 0.68 s,
what should be the magnitude of the minimum deceleration to avoid an accident?
Answer in units of m/s^2

I really need help with this please!

You see I'm not well versed in this subject, and am having a hard time grasping it, but do I just solve the equation 25t + 1/2 at^2 = 170 - 17 and that's it?

To solve this problem, we need to calculate the minimum deceleration the locomotive must have in order to avoid a collision with the car.

Let's break down the problem step by step:

1. Determine the time it takes for the engineer to react:
The reaction time given is 0.68 seconds. This means that once the engineer sees the car, it takes them 0.68 seconds to react.

2. Calculate the distance the train travels during the reaction time:
The speed of the locomotive is given as 25 m/s. In 0.68 seconds, the locomotive would have traveled a distance of (25 m/s) x (0.68 s) = 17 m.

3. Calculate the remaining distance the train needs to stop:
Since the locomotive is 170 m from the crossing when the engineer first sees the car, the remaining distance to stop is (170 m - 17 m) = 153 m.

4. Calculate the time it takes for the locomotive to stop:
The initial velocity is 25 m/s, and the final velocity is 0 m/s (since the train needs to stop). The deceleration (a) is what we need to find. We can use the formula v^2 = u^2 + 2as, where:
- v is the final velocity,
- u is the initial velocity,
- a is the acceleration (deceleration in this case),
- s is the distance.

Rearranging the formula for acceleration, we get a = (v^2 - u^2) / (2s).
Plugging in the values, we have a = (0^2 - 25^2) / (2 * 153).

5. Calculate the minimum deceleration:
Using a calculator, simplifying the equation, we have a = (-625) / 306 ≈ -2.04 m/s^2.

Therefore, the magnitude of the minimum deceleration needed to avoid an accident is approximately 2.04 m/s^2. Note that the negative sign indicates deceleration, or the force opposing the motion of the train.

during the first 0.68s, the train moves 0.68*25 = 17m

Now just solve for a in
25 + at = 0
25t + 1/2 at^2 = 170 - 17