Home » mcq » Civil engineering » Applied mechanics and graphic statics » The equation of motion of a particle starting from rest along a straight line is x = t3 – 3t2 + 5. The ratio of the accelerations after 5 sec and 3 sec will be A. 2 B. 3 C. 4 D. 5
The equation of motion of a particle starting from rest along a straight line is x = t3 – 3t2 + 5. The ratio of the accelerations after 5 sec and 3 sec will be A. 2 B. 3 C. 4 D. 5
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ratio of the velocities after 5 seconds and 3 seconds. The velocity of a particle is given by $v = \frac{dx}{dt}$. In this case, we have $x = t^3 – 3t^2 + 5$, so $v = 3t^2 – 6t$. The velocity after 5 seconds is $v(5) = 3(5)^2 – 6(5) = 25$. The velocity after 3 seconds is $v(3) = 3(3)^2 – 6(3) = 9$. Therefore, the ratio of the velocities is $\frac{v(5)}{v(3)} = \frac{25}{9} \neq 2$.
Option B: $3$. This is the ratio of the positions after 5 seconds and 3 seconds. The position of a particle is given by $x = t^3 – 3t^2 + 5$. In this case, we have $x = t^3 – 3t^2 + 5$, so $x(5) = 5^3 – 3(5)^2 + 5 = 75$. The position after 3 seconds is $x(3) = 3^3 – 3(3)^2 + 5 = 14$. Therefore, the ratio of the positions is $\frac{x(5)}{x(3)} = \frac{75}{14} \neq 3$.
Option C: $4$. This is the ratio of the accelerations after 5 seconds and 3 seconds. As shown above, the acceleration after 5 seconds is $a(5) = 15$ and the acceleration after 3 seconds is $a(3) = 9$. Therefore, the ratio of the accelerations is $\frac{a(5)}{a(3)} = \frac{15}{9} = 5$.
Option D: $5$. This is the ratio of the velocities after 5 seconds and 3 seconds. As shown above, the velocity after 5 seconds is $v(5) = 25$ and the velocity after 3 seconds is $v(3) = 9$. Therefore, the ratio of the velocities is $\frac{v(5)}{v(3)} = \frac{25}{9} \neq 5$.