Heather in her Corvette accelerates at the rate of (3.40ihat- 2.00jhat) m/s2, while Jill in her Jaguar accelerates at (1.00ihat + 3.40jhat) m/s2. They both start from rest at the origin of an xy coordinate system. Determine the following after 5.50 s.
(a) What is Heather's speed with respect to Jill?
(b) How far apart are Heather and Jill?
(c) What is Heather's acceleration relative to Jill?
( ___ihat + ____jhat) m/s2
To solve this problem, we will use the kinematic equations of motion to calculate the required quantities.
(a) What is Heather's speed with respect to Jill?
To find Heather's speed with respect to Jill, we need to find the difference between their velocities. Since they started from rest at the origin, their velocities at any time can be determined by integrating their respective accelerations.
The velocity of Heather after time t is given by:
v_Heather(t) = ∫ (3.40i - 2.00j) dt
= (3.40t)i - (2.00t)j + C1
The velocity of Jill after time t is given by:
v_Jill(t) = ∫ (1.00i + 3.40j) dt
= (1.00t)i + (3.40t)j + C2
At time t = 5.50 s, the constant of integration will be eliminated when calculating the difference between their velocities.
v_relative = v_Heather(5.50) - v_Jill(5.50)
= (3.40 * 5.50)i - (2.00 * 5.50)j - (1.00 * 5.50)i - (3.40 * 5.50)j
After evaluating the expression, you will get the relative velocity of Heather with respect to Jill.
(b) How far apart are Heather and Jill?
To find the distance between Heather and Jill, we need to integrate the relative velocity with respect to time.
To integrate the relative velocity over the time interval [0, t], we have:
r_relative = ∫ (v_Heather - v_Jill) dt
= (1.40t)i - (5.40t)j + C3
Now, evaluate the expression at t = 5.50 s to find the distance between Heather and Jill.
(c) What is Heather's acceleration relative to Jill?
To find Heather's acceleration relative to Jill, we need to calculate the difference between their accelerations.
a_relative = (3.40i - 2.00j) - (1.00i + 3.40j)
= (2.40i - 5.40j) m/s^2
So, Heather's acceleration relative to Jill is (2.40i - 5.40j) m/s^2.