Question

A particle moves in a straight line such that its velocity, v ms, at time t seconds is given by

v=4t² - 6t + 9 - 2 sin(4t), 0≤t≤1.
The particle's acceleration is zero at t = T.

Answer 1

The student's question involves finding the time at which a particle's acceleration, which can be determined by differentiating the given velocity function, equals zero. Calculating the acceleration function from the given velocity function will allow them to solve for the time when acceleration is zero.

Step-by-step explanation:

The question pertains to the calculation of acceleration and velocity in one-dimensional motion given a velocity-time function. The specific instance where the particle's acceleration is zero at time t = T requires the student to find the time T when this condition is met. This involves differentiating the given velocity function to find the acceleration function and then solving for T. To calculate acceleration, differentiate the given velocity function with respect to time:

v(t) = 4t² - 6t + 9 - 2 sin(4t)

a(t) = dv/dt = d(4t² - 6t + 9 - 2 sin(4t))/dt

a(t) = 8t - 6 - 8 cos(4t)

To find the time T when acceleration is zero:

0 = 8T - 6 - 8 cos(4T)

This equation must be solved to find the value of T that satisfies the condition a(T)=0. Since this could involve transcendent equation solving, the value of T may require numerical methods or graphing to determine.