A particle's motion is described by x(t)=8t-2t2. (where x is in meters and t is in seconds)
f) How fast is the particle moving at t=3s?
______m/s
g) How much is it acceleration at t=3s?
______m/s2
v=dx/dt =8-4t = 8 - 4•3= = -4 m/s.
a=dv/dt== - 4 m/s²
To find the answers for these questions, we need to take the derivative of the position function x(t) with respect to time, t. The first derivative of x(t) will give us the velocity function, and the second derivative will give us the acceleration function.
Let's start with finding the velocity function at t=3s (question f).
Step 1: Find the first derivative of x(t) with respect to t.
x(t) = 8t - 2t^2
To find dx/dt (the derivative of x with respect to t), we can use the power rule for differentiation:
dx/dt = 8 - 4t
Step 2: Substitute t=3s into the velocity function.
v(t) = dx/dt = 8 - 4t
Substituting t=3 into the velocity function:
v(3) = 8 - 4(3)
v(3) = 8 - 12
v(3) = -4 m/s
Therefore, the particle's speed at t=3s is -4 m/s (to indicate that it is moving in the negative direction).
Moving on to question g, we need to find the acceleration at t=3s.
Step 3: Find the second derivative of x(t) with respect to t.
Taking the derivative of v(t) = 8 - 4t, we get:
dv/dt = -4
This is the acceleration function.
Step 4: Substitute t=3s into the acceleration function.
a(t) = dv/dt = -4
Substituting t=3 into the acceleration function:
a(3) = -4 m/s^2
Therefore, the particle's acceleration at t=3s is -4 m/s^2.