Question

h(x)=3x^2+2Kx-3 ; x= -1/2​

Answer 1

$Solution Distance between two points (s) =\bf{180 m}180m Time taken to cover the distance (t) = \bf{6 s}6s Final velocity (v) = \bf{45 m/s}45m/s Finding the initial velocity and acceleration:- We know:----- The 1st equation of motion is:- \boxed{\bf{v \: = \: u \: + \: at}} v=u+at Hence, {45 \: = \: u \: + 6a}45=u+6a {u = 45 - 6a}\: \: \: \longrightarrow {\bf(eqn.1)}u=45−6a⟶(eqn.1) Using the 2nd equation of motion:- We have:- \boxed{\bf{s = ut + \frac{1}{2} \: a {t}^{2}}} \: \: \: \longrightarrow {\bf(eqn.2)} s=ut+ 2 1 at 2 ⟶(eqn.2) Substituting (eq .1) in (eq .2):- 180 = (45 - 6a)6 + ( \frac{1}{2} a \times 36)( 2 1 a×36) 180 = 270 - 36a + \frac{36a}{2} 2 36a 180 = 270 - 36a(1 - \frac{1}{2})(1− 2 1 ) 180 = 270 - 36( \frac{1}{2} )( 2 1 ) 180 = 270 - 18a 18a = 270 - 180 18a = 90 a = \frac{90}{18} 18 90 a = {\cancel{\frac{90}{18}}} 18 90 \implies \bf{5 \: m {s}^{ - 2}}⟹5ms −2 Putting, the value of acceleration [a] in equation (1):- We have:- u = 45 - (6 \times 5)(6×5) u = 45 - 30 \bf{u = 15 \: m {s}^{ - 1}}u=15ms −1 Therefore, \large \bf{Initial \: velocity[u] = 15 m/s^{1}}Initialvelocity[u]=15m/s 1 \large \bf{acceleration[a]= 5m/s^{2}}acceleration[a]=5m/s 2 ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬SolutionDistancebetweentwopoints(s)=180m180mTimetakentocoverthedistance(t)=6s6sFinalvelocity(v)=45m/s45m/sFindingtheinitialvelocityandacceleration:−Weknow:−−−−−The1stequationofmotionis:−v=u+atv=u+atHence,45=u+6a45=u+6au=45−6a⟶(eqn.1)u=45−6a⟶(eqn.1)Usingthe2ndequationofmotion:−Wehave:−s=ut+21at2⟶(eqn.2)s=ut+21at2⟶(eqn.2)Substituting(eq.1)in(eq.2):−180=(45−6a)6+(21a×36)(21a×36)180=270−36a+236a236a180=270−36a(1−21)(1−21)180=270−36(21)(21)180=270−18a18a=270−18018a=90a=18901890a=18901890⟹5ms−2⟹5ms−2Putting,thevalueofacceleration[a]inequation(1):−Wehave:−u=45−(6×5)(6×5)u=45−30u=15ms−1u=15ms−1Therefore,Initialvelocity[u]=15m/s1Initialvelocity[u]=15m/s1acceleration[a]=5m/s2acceleration[a]=5m/s2▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬$