Maths Made Easy Forums Standard Formulae & Equations Stopping Distance

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Keymaster
Post count: 659

Sarah is travelling at 75km/h and suddenly has to stop. Her reaction time is 4.2 seconds. She comes to a stop 116.19m further down the road. Use the braking distance formula d = kv2 to find how much further, to the nearest metre, her stopping distance would be if she was travelling 12km/h faster.

Stopping distance = braking distance + reaction time distance

Find the reaction time distance

75km/h change to m/s by multiplying  by 1000 to change km to m and dividing by 60 and then divide by 60 again to change from hours to minutes to seconds

speed in m/s = 75 × 1000 ÷ 60 ÷ 60 = 20.83m/s

reaction time = 4.2s
reaction time distance = 20.83 × 4.2 = 87.5m

116.19 = braking distance + 87.5

braking distance = 116.19 -87.5 = 28.69m

substitute this distance into the formula to find k

28.69 =  k × 752

$k=\frac{28.69}{5625}$

k = 0.0051

∴ d = 0.0051v2

12km/h faster = 75 + 12 = 87

find braking distance

d = 0.0051 × 872

d = 38.6m

find reaction distance

speed in m/s = 87 × 1000 ÷ 60 ÷ 60 = 24.16m/s

reaction time distance = 24.16 × 4.2 = 101.5m

stopping distance = 38.6 + 101.5 = 140.1m

difference is 140.1 – 116.19

= 23.91m

Keymaster
Post count: 659

Raine is driving at a speed of 80 km/h. It takes Raine two seconds to react to a dangerous situation before applying the brakes. The stopping distance (in m) is given by the formula: Stopping distance: $d=\frac{5Vt}{18}+\frac{V^2}{170}$. How far will Raine travel in her car after applying the brakes using this formula?

Substitute V = 80 and t = 2
$d=\frac{5\times80\times2}{18}+\frac{80^2}{170}$
= 82.0915
= 82m