Question

A train moving at 36kmh-1 is brought to rest in 10s by applying brakes. Find a. The retardation produced by brakes b. Distance covered by the train before it stops​

Answer 1

Explanation:

First, let's translate this into smaller units, shall we? There are 1000 meters in a kilometer, and 3600 seconds per hour. So, multiply thus:

36km/h × 1000m/km / 3600s/h = 10m/s This is your initial speed, or v₀. We already know the final speed, or vₓ.

The negative acceleration is given by

a = (vₓ - v₀) / Δt

where Δt is the amount of elapsed time. Plugging in,

a = (0m/s - 10m/s) / 10s = -1m/s²

or, plugging back in the original units, -3.6km/h/s

The distance travelled is the average velocity times elapsed time (it must be an absolute value because distance is never negative). So,

d = |((vₓ - v₀) / 2) × Δt| = |((0m/s - 10m/s) / 2) × 10s| = 50m.

Answer 2

Explanation:

First, let's translate this into smaller units, shall we? There are 1000 meters in a kilometer, and 3600 seconds per hour. So, multiply thus:

36km/h × 1000m/km / 3600s/h = 10m/s This is your initial speed, or v₀. We already know the final speed, or vₓ.

The negative acceleration is given by

a = (vₓ - v₀) / Δt

where Δt is the amount of elapsed time. Plugging in,

a = (0m/s - 10m/s) / 10s = -1m/s²

or, plugging back in the original units, -3.6km/h/s

The distance travelled is the average velocity times elapsed time (it must be an absolute value because distance is never negative). So,

d = |((vₓ - v₀) / 2) × Δt| = |((0m/s - 10m/s) / 2) × 10s| = 50m.