The distance of a particle traveling on a straight line from point A is $$s=10t^2+t^3$$. The rate of change of acceleration at time $$t=4$$ is:
The distance of a particle traveling on a straight line from point A is $$s=10t^2+t^3$$. The rate of change of acceleration at time $$t=4$$ is:
1 Answer
James Dowd
Updated on January 2nd, 2021Acceleration is simply the second derivative of the expression for distance. Therefore, to solve this problem, we want to first take the second derivative of the expression:
Since we have now found the second derivative, we know:
In order to find our acceleration at t=4, we simply plug this t value into our expression for acceleration:
Therefore, our acceleration is equal to 44 at t=4.
Related Questions
The position of a particle that traverses a straight line is expressed as $$x(t)=t^3+2t^2-t+10$$, where x is in feet and t in seconds. What is the acceleration (ft/$$sec^2$$) of the particle when the velocity is zero?
A tractor with a mass of 2,300 kg is traveling at 10 m/s. At time t = 0, the driver pulls the emergency brake and the tractor begins to slide. The properties of the relevant materials, are shown:
Material 1 Material 2 Static Friction Kinetic Friction Rubber Asphalt 0.9 0.4 At t = 3, the speed (m/s) of the tractor is most nearly:
A 10-kg block which starts at rest, begins sliding after being pushed by a constant force F of 15N. The time in seconds it takes for the block to reach 20 m/s is most nearly.
A sea vessel accelerates at a rate of $$22\ ft//sec^2$$. The vessel travels 140 ft while its speed changes to 70 ft/sec. The initial velocity (ft/sec) was most nearly: