Motion In A Plane With Constant Acceleration

Let the acceleration of car P be , which increases linearly with time. Let the acceleration of car Q be constant . The velocity of car P at time is given by: The velocity of car Q is given by: At , both cars are at the same position. As time progresses, the velocities of both cars change, and they may cross each other again. Case I:
- increases quadratically, while increases linearly.
- The cars will cross twice, once when and again when the positions of the cars match at a later time.
Case II:
- For a constant acceleration , car P crosses car Q when the difference in their velocities leads to the second crossing.
Thus, the total number of crossings is 3, including the crossing at .

Concept: Acceleration is given by Using the chain rule:
Step 1:Identify velocity components}
Step 2:Compute acceleration components} Thus
Step 3:Substitute the point }
Step 4:Find magnitude} Thus the magnitude becomes

Step 1: Understanding the Concept:
By utilizing Newton's second law, we can determine the 2D acceleration vector of the mass. Since the force is constant and the body starts from rest, we can use the kinematic equations for uniform acceleration to find its position vector at s and compare it with the given coordinate expressions.

Step 2: Key Formula or Approach:
Newton's Second Law:
Kinematics (from rest, ):

Step 3: Detailed Explanation:
Mass .
Force .
Calculate the acceleration vector:
.
The body starts from rest, so initial velocity .
The position vector after s is:


.
We are given the position after 2 seconds as m.
Equating the components:
x-component: .
y-component: .
Find the ratio :
.

Step 4: Final Answer:
The ratio is 5 : 4.