When the train running at 36km / h starts to go downhill, the acceleration on the slope is equal to 0.2m/s2, and reaches the bottom of the slope after 30s. The length of the slope and the speed when the train reaches the bottom of the slope are calculated

When the train running at 36km / h starts to go downhill, the acceleration on the slope is equal to 0.2m/s2, and reaches the bottom of the slope after 30s. The length of the slope and the speed when the train reaches the bottom of the slope are calculated

According to the displacement formula, x = v0t + 12at2 = 10 × 30 + 12 × 0.2 × 302M = 390m; according to the speed formula, v = V0 + at = 10 + 0.2 × 30m / S = 16m / s

The train running at 36km / h starts to go downhill, and the acceleration on the slope is equal to 0.2m/s2. After 30s, the train reaches the bottom of the slope, and the length of the slope is calculated
Is it calculated by x = VO * t + (1 / 2) at ^ 2
Why not use x = (VO + V) * t / 2

Of course, it can be used. It's essentially the same
v0=36km/h=10m/s
v=10+0.2*30=16m/s
x=(10+16)*30/2=390m
x=1/2*0.2*30^2+10*30=390m
The results obtained by the two methods are the same

When the train running at 36km / h starts to go downhill, the acceleration on the slope is equal to 0.2m/s2, and reaches the bottom of the slope after 30s. The length of the slope and the speed when the train reaches the bottom of the slope are calculated

According to the displacement formula, x = v0t + 12at2 = 10 × 30 + 12 × 0.2 × 302M = 390m; according to the speed formula, v = V0 + at = 10 + 0.2 × 30m / S = 16m / s