To determine the velocity of the particle when its acceleration becomes zero, we first need to understand the relationship between position, velocity, and acceleration for this motion.
\(v = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 20t + 15)\)
Calculating the derivative, we get: \(v = 3t^2 - 12t + 20\)
\(a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 20)\)
Calculating the derivative, we get: \(a = 6t - 12\)
\(6t - 12 = 0\)
Solving for \(t\), we get: \(6t = 12 \quad \Rightarrow \quad t = 2\)
\(v = 3(2)^2 - 12(2) + 20\)
\(v = 3(4) - 24 + 20\)
\(v = 12 - 24 + 20\)
\(v = 8 \, \text{m/s}\)
Hence, the correct answer is 8 m/s.
Given the position function:
\(x = t^3 - 6t^2 + 20t + 15\)
Step 1: Find the velocity \( v \):
The velocity is the first derivative of the position function with respect to time:
\(v = \frac{dx}{dt} = 3t^2 - 12t + 20\)
Step 2: Find the acceleration \( a \):
The acceleration is the derivative of velocity:
\(a = \frac{dv}{dt} = 6t - 12\)
Step 3: When is the acceleration zero?
Set the acceleration to zero to find the time at which the acceleration becomes zero:
\(6t - 12 = 0 \implies t = 2 \, \text{sec}\)
Step 4: Find the velocity at \( t = 2 \):
Substitute \( t = 2 \) into the velocity equation:
\(v = 3(2)^2 - 12(2) + 20 = 3(4) - 24 + 20 = 12 - 24 + 20 = 8 \, \text{m/s}\)
Thus, the velocity of the body when its acceleration becomes zero is. \( 8 \, \text{m/s} \)
The Correct Answer is: \( 8 \, \text{m/s} \)
A laser beam has intensity of $4.0\times10^{14}\ \text{W/m}^2$. The amplitude of magnetic field associated with the beam is ______ T. (Take $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2/\text{N m}^2$ and $c=3\times10^8\ \text{m/s}$)
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
