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Question Number 136353 by BHOOPENDRA last updated on 21/Mar/21

Commented by BHOOPENDRA last updated on 21/Mar/21

$m r . W s i r h e r e i s f u l l q u e s t i o n ?$
	Answered by mr W last updated on 21/Mar/21

	Commented by mr W last updated on 21/Mar/21
	$Part I a(t)= { ((0 for 0≤t<t_1 )),((−c(t−t_1 ) for t_1 ≤t≤t_2 )) :} 0≤t<t_1 : v(t)=v_0 s(t)=v_0 t s(t_1 )=v_0 t_1 t_1 ≤t≤t_2 : v(t)=v_0 +∫_t_1 ^t a(t)dt =v_0 −c∫_t_1 ^t (t−t_1 )dt =v_0 −c[(((t−t_1 )^2 )/2)]_t_1 ^t =v_0 −((c(t−t_1 )^2 )/2) v(t_2 )=v_0 −((c(t_2 −t_1 )^2 )/2)=0 ⇒v_0 =((c(t_2 −t_1 )^2 )/2) s(t)=s(t_1 )+∫_t_1 ^t v(t)dt =v_0 t_1 +∫_t_1 ^t [v_0 −((c(t−t_1 )^2 )/2)]dt =v_0 t_1 +[v_0 t−((c(t−t_1 )^3 )/6)]_t_1 ^t =v_0 t_1 +[v_0 t−((c(t−t_1 )^3 )/6)−v_0 t_1 ] =v_0 t−((c(t−t_1 )^3 )/6)$
	$P a r t I$ $a (t) = {\begin{cases} 0 f o r 0 ⩽ t < t_{1} \\ - c (t - t_{1}) f o r t_{1} ⩽ t ⩽ t_{2} \end{cases}$ $0 ⩽ t < t_{1} :$ $v (t) = v_{0}$ $s (t) = v_{0} t$ $s (t_{1}) = v_{0} t_{1}$ $t_{1} ⩽ t ⩽ t_{2} :$ $v (t) = v_{0} + \int_{t_{1}}^{t} a (t) d t$ $= v_{0} - c \int_{t_{1}}^{t} (t - t_{1}) d t$ $= v_{0} - c {[\frac{{(t - t_{1})}^{2}}{2}]}_{t_{1}}^{t}$ $= v_{0} - \frac{c {(t - t_{1})}^{2}}{2}$ $v (t_{2}) = v_{0} - \frac{c {(t_{2} - t_{1})}^{2}}{2} = 0$ $\Rightarrow v_{0} = \frac{c {(t_{2} - t_{1})}^{2}}{2}$ $s (t) = s (t_{1}) + \int_{t_{1}}^{t} v (t) d t$ $= v_{0} t_{1} + \int_{t_{1}}^{t} [v_{0} - \frac{c {(t - t_{1})}^{2}}{2}] d t$ $= v_{0} t_{1} + {[v_{0} t - \frac{c {(t - t_{1})}^{3}}{6}]}_{t_{1}}^{t}$ $= v_{0} t_{1} + [v_{0} t - \frac{c {(t - t_{1})}^{3}}{6} - v_{0} t_{1}]$ $= v_{0} t - \frac{c {(t - t_{1})}^{3}}{6}$
	Commented by mr W last updated on 21/Mar/21

	Commented by BHOOPENDRA last updated on 21/Mar/21

	$t h a n k u s i r$
	Commented by BHOOPENDRA last updated on 21/Mar/21

	$p a r t b s i r ?$
	Commented by mr W last updated on 21/Mar/21

	$P a r t I I$ $g i v e n c = 6 m / s^{3}, t_{1} = 1 s, v_{0} = 12 m / s^{2}$ $v_{0} = \frac{c {(t_{2} - t_{1})}^{2}}{2} = \frac{6 {(t_{2} - 1)}^{2}}{2} = 12$ $\Rightarrow t_{2} = 3 s$ $s (t_{2}) = v_{0} t_{2} - \frac{c {(t_{2} - t_{1})}^{3}}{6}$ $= 12 \times 3 - \frac{6 {(3 - 1)}^{3}}{6} = 28 m$ $v e l o c i t y o f b i k e r :$ $v_{B} = \frac{17 + 28}{3} = 15 m / s$
	Commented by BHOOPENDRA last updated on 21/Mar/21

	$t h a n k s s i r$
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