A car stops with an acceleration of -20 m/s/s in 8 seconds. How far did it go while stopping?

640 m.

Explanation:

The following data were obtained from the question:

Acceleration (a) = –20 m/s/s

Time (t) = 8 s

Final velocity (v) = 0 m/s

Distance (s) =.?

Next, we shall determine the initial velocity (u) of the car. This can be obtained as follow:

Acceleration (a) = –20 m/s/s

Time (t) = 8 s

Final velocity (v) = 0 m/s

Initial velocity (u)

a = (v – u) / t

–20 = (0 – u) / 8

–20 = – u / 8

Cross multiply

–20 × 8 = – u

– 160 = – u

Divide both side by – 1

u = – 160 / – 1

u = 160 m/s

Finally, we shall determine the distance travelled by the car before stopping as follow:

Time (t) = 8 s

Final velocity (v) = 0 m/s

Initial velocity (u) = 160 m/s

Distance (s) =.?

s = (v + u)t /2

s = (0 + 160) × 8 /2

s = (160 × 8) /2

s = 1280 / 2

s = 640 m

Therefore, the car travelled 640 m before stopping.

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(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

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