Question

Two cars $P$ and $Q$ start from a point at the same time in a straight line and their positions are represented by $x _{ P }( t )= at + bt ^{2}$ and $x _{ Q }( t )=f t - t ^{2}$. At what time do the cars have the same velocity ?

• A. $\frac{a - f}{1 + b}$
• B. $\frac{a + f}{2( b - 1)}$
• C. $\frac{a + f}{2( 1 + b)}$
• D. $\frac{f - a }{2( 1 + b)}$

Answer 1

The Correct Option is D

To find the time when the two cars, P and Q, have the same velocity, we need to first determine their respective velocity functions and then set them equal to one another.

### Step 1: Determine the Velocity Functions

• The position of car P is given by x_{P}(t) = at + bt^{2}. The velocity is obtained by differentiating the position with respect to time: v_{P}(t) = \frac{d}{dt}(at + bt^{2}) = a + 2bt.
• The position of car Q is given by x_{Q}(t) = ft - t^{2}. The velocity is obtained similarly: v_{Q}(t) = \frac{d}{dt}(ft - t^{2}) = f - 2t.

### Step 2: Equate the Velocities

• To find the time when their velocities are equal, set the two velocity functions equal: a + 2bt = f - 2t.

### Step 3: Solve for t

• Rearrange and solve for t:
  • Bring all the terms involving t to one side: 2bt + 2t = f - a.
  • Factor out t: t(2b + 2) = f - a.
  • Solve for t: t = \frac{f - a}{2(1 + b)}.

Therefore, the cars have the same velocity at time t = \frac{f - a}{2(1 + b)}.