The driver of a train moving at 23m/s applies the breaks when it pases an amber signal. The next signal is 1km down the track and the train reaches it 76s later. The acceleration is - 0.26s^2. Find its speed at the next signal.

3.2 m/s

Explanation:

Given:

Δx = 1000 m

v₀ = 23 m/s

a = - 0.26 m/s²

t = 76 s

Find: v

This problem is over-defined. We only need 3 pieces of information, and we're given 4. There are several equations we can use. For example:

v = at + v₀

v = (-0.26 m/s²) (76 s) + (23 m/s)

v = 3.2 m/s

Or:

Δx = ½ (v + v₀) t

(1000 m) = ½ (v + 23 m/s) (76 s)

v = 3.3 m/s

Or:

v² = v₀² + 2aΔx

v² = (23 m/s) ² + 2 (-0.26 m/s²) (1000 m)

v = 3.0 m/s

Or:

Δx = vt - ½ at²

(1000 m) = v (76 s) - ½ (-0.26 m/s²) (76 s) ²

v = 3.3 m/s

As you can see, you get slightly different answers depending on which variables you use. Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.