A vehicle approaches a set of traffic lights with a speed of 15m/s. When 50m from the lights the driver brakes and comes to a rest at the lights. Calculate the deceleration of the vehicle

–2.25 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 15 m/s

Distance travelled (s) = 50 m

Final velocity (v) = 0 m/s

Deceleration (a) =?

v² = u² + 2as

0² = 15² + (2 × a × 50)

0 = 225 + 100a

Collect like terms

0 – 225 = 100a

– 225 = 100a

Divide both side by 100

a = –225/100

a = –2.25 m/s²

Thus, the deceleration of the vehicle is –2.25 m/s²

2b2t

explanation:

2b2t

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer