Question

When A overtakes B.

• A. The time when A overtakes B is $\frac{5 + \sqrt{105}}{2}$ seconds.
• B. The time when A overtakes B is $\frac{5 - \sqrt{105}}{2}$ seconds.
• C. The time when A overtakes B is $5$ seconds.
• D. The time when A overtakes B is $10$ seconds.

Answer 1

Answer: (A)

The time when A overtakes B is $\frac{5 + \sqrt{105}}{2}$ seconds.

Answer 2

Let $x_A(t)$ and $x_B(t)$ be the positions of objects A and B at time $t$. We set the initial position of object A as the origin, so $x_{A0} = 0$. Given:

• Initial velocity of A: $v_{A0} = 5 \, m/s$
• Acceleration of A: $a_A = 2 \, m/s^2$
• Initial position of B: $x_{B0} = 20 \, m$
• Velocity of B: $v_B = 10 \, m/s$ (constant)

The position of object A at time $t$ is given by: $x_A(t) = x_{A0} + v_{A0}t + \frac{1}{2}a_A t^2$ $x_A(t) = 0 + 5t + \frac{1}{2}(2)t^2$ $x_A(t) = 5t + t^2$

The position of object B at time $t$ is given by: $x_B(t) = x_{B0} + v_B t$ $x_B(t) = 20 + 10t$

Object A overtakes object B when their positions are equal, i.e., $x_A(t) = x_B(t)$. $5t + t^2 = 20 + 10t$

Rearranging the terms to form a quadratic equation: $t^2 - 5t - 20 = 0$

Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-20)}}{2(1)}$ $t = \frac{5 \pm \sqrt{25 + 80}}{2}$ $t = \frac{5 \pm \sqrt{105}}{2}$

Since time must be positive, we take the positive root: $t = \frac{5 + \sqrt{105}}{2}$ seconds.