Question

Q3. A car moving with a speed of 36km/hr is brought to rest in 10s. Calculate the retardation and the distance moved by the car before coming to rest.​

Answer 1

Answer:

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

Initial velocity (u) = 36km/hr = 10 m/s
Final velocity (v) = 0 m/s
Time (t) = 10s
Retardation = v-u/t
= 0 - 10 / 10
= -1 m/s2

Distance = ut + 1/2 at 2
= 10*10 + 1/2 (-1)
= 100+1/2*(-1)
= 100/2
= 50 m