Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

An object travels along a straight line. The function v(t)=4t3v(t) = 4\sqrt{t} - 3 gives the object's velocity, in kilometers per hour, at time t > 0 hours.\newlineWrite a function that gives the object's acceleration a(t)a(t) in kilometers per hour per hour.\newlinea(t) = ______

Full solution

Q. An object travels along a straight line. The function v(t)=4t3v(t) = 4\sqrt{t} - 3 gives the object's velocity, in kilometers per hour, at time t>0t > 0 hours.\newlineWrite a function that gives the object's acceleration a(t)a(t) in kilometers per hour per hour.\newlinea(t) = ______
  1. Differentiate Velocity Function: To find the acceleration function a(t)a(t), we need to differentiate the velocity function v(t)v(t) with respect to time tt. The velocity function given is v(t)=4t3v(t) = 4\sqrt{t} - 3.
  2. Apply Power Rule: Differentiate v(t)=4t3v(t) = 4\sqrt{t} - 3. Using the power rule, t\sqrt{t} can be rewritten as t(1/2)t^{(1/2)}. So, the derivative of t(1/2)t^{(1/2)} is (1/2)t(1/2)(1/2)t^{(-1/2)}. Applying the constant multiple rule, the derivative of 4t(1/2)4t^{(1/2)} is 4×(1/2)t(1/2)=2t(1/2)4 \times (1/2)t^{(-1/2)} = 2t^{(-1/2)}. The derivative of a constant (3)(-3) is 00.
  3. Acceleration Function Calculation: Therefore, the acceleration function a(t)=2t(1/2)a(t) = 2t^{(-1/2)}, which simplifies to 2t\frac{2}{\sqrt{t}}.

More problems from Relate position, velocity, speed, and acceleration using derivatives