A particle is projected along an inclined plane with a velocity given by the formula
v = 4.4t2 + 10.1t -23.2. Determine the acceleration of the particle when t = 1.5 s.
acceleration in m/s2 =
My Answer
V'=8.8t+10.1
V"=8.8
V"(1.5)= 8.8
acceleration in m/s2 = 8.8
Can you please confirm this is correct.
Thanks
To determine the acceleration of the particle, you need to take the derivative of the velocity equation with respect to time. Let's do that step by step:
Step 1: Differentiate the velocity equation.
v = 4.4t^2 + 10.1t - 23.2
To differentiate the equation, we need to apply the power rule. The power rule states that if we have a term of the form a*t^n, where a and n are constants, then the derivative is n*a*t^(n-1).
Differentiating each term:
dv/dt = 2*4.4t^(2-1) + 1*10.1t^(1-1) + 0
=8.8t + 10.1
Therefore, the derivative of the velocity equation is dv/dt = 8.8t + 10.1.
Step 2: Evaluate the derivative at t = 1.5s.
To find the acceleration at a specific time, substitute the given time value into the derivative equation.
dv/dt = 8.8t + 10.1
dv/dt = 8.8(1.5) + 10.1
= 13.2 + 10.1
= 23.3 m/s^2
Thus, the acceleration of the particle when t = 1.5s is 23.3 m/s^2.
Therefore, your previous answer that the acceleration is 8.8 m/s^2 is incorrect. The correct answer is 23.3 m/s^2.