Question

Two cars P and Q are moving on a road in the same direction. Acceleration of car P increases linearly with time whereas car Q moves with a constant acceleration. Both cars cross each other at time \( t = 0 \), for the first time. The maximum possible number of crossing(s) (including the crossing at \( t = 0 \)) is:

Hint:

In such problems, consider the velocity-time relationship for both cars and solve for when their positions are equal to find the crossing points.

Answer 1

Correct Answer: 3

Let the acceleration of car P be \( a_P = k t \), which increases linearly with time. Let the acceleration of car Q be constant \( a_Q = a \). The velocity of car P at time \( t \) is given by: \[ v_P = \int a_P \, dt = \int k t \, dt = \frac{k t^2}{2} \] The velocity of car Q is given by: \[ v_Q = \int a_Q \, dt = a t \] At \( t = 0 \), 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:
- \( v_P \) increases quadratically, while \( v_Q \) increases linearly.
- The cars will cross twice, once when \( t = 0 \) and again when the positions of the cars match at a later time.
Case II:
- For a constant acceleration \( a_Q = a \), 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 \( t = 0 \).

Answer 2

Step 1: Understand the situation.
Two cars, P and Q, are moving in the same direction on a straight road.
- Car P has an acceleration that increases linearly with time, i.e., \( a_P = kt \) where \( k \) is a constant.
- Car Q moves with a constant acceleration, say \( a_Q = c \).
Both cars cross each other for the first time at \( t = 0 \). We need to find the maximum possible number of crossings (including the one at \( t = 0 \)).

Step 2: Write the equations of motion.
Let the initial velocities of the two cars at \( t = 0 \) be \( u_P \) and \( u_Q \).
For car P:
\[ a_P = kt \Rightarrow v_P = u_P + \int a_P \, dt = u_P + \frac{1}{2}kt^2 \] \[ x_P = \int v_P \, dt = u_Pt + \frac{1}{6}kt^3 \]
For car Q:
\[ a_Q = c \Rightarrow v_Q = u_Q + ct \] \[ x_Q = \int v_Q \, dt = u_Qt + \frac{1}{2}ct^2 \]

Step 3: Find the condition for crossing.
Both cars cross when their displacements are equal, i.e., \( x_P = x_Q \):
\[ u_Pt + \frac{1}{6}kt^3 = u_Qt + \frac{1}{2}ct^2 \] Simplify the equation:
\[ \frac{1}{6}kt^3 - \frac{1}{2}ct^2 + (u_P - u_Q)t = 0 \] \[ t\left(\frac{1}{6}kt^2 - \frac{1}{2}ct + (u_P - u_Q)\right) = 0 \] We already know one crossing at \( t = 0 \). The remaining crossings depend on the quadratic factor:
\[ \frac{1}{6}kt^2 - \frac{1}{2}ct + (u_P - u_Q) = 0 \]

Step 4: Find the number of crossings.
This is a quadratic equation in \( t \). The number of additional crossings depends on the number of real roots of this quadratic equation.
For real crossings, the discriminant must be positive:
\[ \Delta = \left(-\frac{1}{2}c\right)^2 - 4\left(\frac{1}{6}\right)(u_P - u_Q)k > 0 \] Simplify: \[ \frac{c^2}{4} - \frac{2}{3}k(u_P - u_Q) > 0 \] If this condition is satisfied, there can be two additional real roots (crossings). Including the first crossing at \( t = 0 \), the total number of crossings can be three.

Step 5: Final conclusion.
Hence, the maximum possible number of crossings (including the one at \( t = 0 \)) is:
\[ \boxed{3} \]