1.

When the brakes are applied to moving car, the car travels a distance ‘s’ ft. in ‘t’ see given by s = 8t – 6t2 when does the car stop?

Answer»

Given s = 8t – 6t2 

v = \(\frac{ds}{dt}\)= 8 – 12t car stops when v = 0

∴ 8 – 12t = 0 ⇒ t = \(\frac{8}{12}\)= \(\frac{2}{3}\) sec.



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