Given, v(t) = 2t(3 – t) = 6t – 2t2
(a) For maximum velocity,
\(\frac{dv(t)}{dt}\) = 0
⇒ \(\frac{d}{dt}\) (6t − 2t2 ) = 0
⇒ 6 − 4t = 0⇒ t =\(\frac{6}{4}\) = \(\frac{3}{2}\)s = 1.5 s
(b) \(\frac{ds}{dt}\) = 6t − 2t2 [s = displacement]
ds = (6t – 2t2 ) dt
∴ Distance travelled in time interval to 0 to 3s.
s = \(\int_0^3(6t -2t^2)dt\)
= \({[\frac{6t^2}{2}-\frac{2t^3}{3}]}^3_0={[3t^2-\frac{2}{3}t^3]}^3_0\)
= 3×9− \(\frac{2}{3}\) × 3 × 3 × 3 = 27 – 18 = 9 m.
Average velocity = \(\frac{9}{3}\) = 3 m/s
As, vavg = 6t − 2t2 ⇒ 3 = 6t – 2t2⇒2t2 − 6t − 3 = 0
⇒ \(t=\frac{6\pm\sqrt{6^2-4\times2\times3}}{2\times2}=\frac{6\pm\sqrt{36-24}}{4}=\frac{6\pm\sqrt {12}}{4}=\frac{3\pm\sqrt 3}{2}\)
Taking ‘+’ ve sign we get \(t_1=\frac{3+\sqrt 3}{2}\) = 2.36 sec.
Taking ‘−‘ ve sign we get \(t^2=\frac{3-\sqrt 3}{2}\) = 0.633 sec
which is less than the least of the clock so it cannot be measured,
∴ rejecting t2, the average velocity is maximum at 2.36 sec.
(c) Acceleration will be maximum when velocity is zero
∴ (6t − t2 ) = 0
t(6−2t) = 0
t = 0, 3 sec
(d) \(s_1=\int_0^3(6t - 2t^2)dt\)
= \(\left.\frac{6t^2}{2}-\frac{2t^3}{3}\right|_0^3\)
= 9 m
\(s_2=\int_3^6(-(t-3)(6-t))dt\)
= \(\int_3^6-(6t-t^2-18+3t)dt\)
= \(\int_3^6-(t^2+9t-18)dt\)
= \(\left.\frac{t^3}{3}-\frac{9t^2}{2}+18t\right|_3^6\)
= −4.5 m
Total distance travelled in one cycle is
= 9 – 4.5 m = 4.5 m
No. of cycles = \(\frac{20}{4.5}\) = 4.44 cycles ≈ 5 cycles