AS Level Maths 9709: Two Cars Meeting, suvat and a Repeated Root

Car $A$ starts from rest at a fixed point $O$ on a straight road and moves with constant acceleration $2\text{ m s}^{-2}$ . Car $B$ travels along the same road in the same direction at a constant velocity of $20\text{ m s}^{-1}$ , and passes through $O$ exactly $5$ seconds after Car $A$ sets off. Let $t$ be the time in seconds measured from the instant Car $A$ sets off.

(a) Write down expressions, in terms of $t$ , for the displacement from $O$ of Car $A$ , and of Car $B$ (valid for $t\ge5$ ). [3]

(b) Show that the times at which the two cars are at the same distance from $O$ satisfy $(t-10)^2=0$ , and hence find this value of $t$ . [3]

(c) State what a repeated root tells you about the motion of the two cars, and find the distance from $O$ at which the two cars are momentarily at the same position. [2]

Part (a): Displacement expressions

Car A starts from rest ( $u=0$ ) at $O$ with constant acceleration $a=2\text{ m s}^{-2}$ . Using $s=ut+\tfrac12at^2$ :

$s_A = 0\cdot t + \frac12(2)t^2 = t^2$

Car B travels at constant velocity $20\text{ m s}^{-1}$ , but does not pass $O$ until $t=5$ . For $t\ge5$ , the time Car B has been travelling is $(t-5)$ , so:

$s_B = 20(t-5)$

Part (b): Finding when the cars are level

The cars are at the same distance from $O$ when $s_A=s_B$ (for $t\ge5$ ):

$t^2 = 20(t-5)$

Expanding the right-hand side:

$t^2 = 20t - 100$

Rearranging:

$t^2 - 20t + 100 = 0$

This factorises as a perfect square:

$(t-10)^2 = 0$

as required. This gives a single (repeated) solution:

$t = 10$

Recompute independently as a check: expanding $(t-10)^2 = t^2-20t+100$ , which matches the equation above exactly. Substituting $t=10$ back into the original equation: $10^2=100$ on the left, and $20(10-5)=20(5)=100$ on the right. Both sides agree, confirming $t=10$ is correct.

Part (c): Interpreting the repeated root

A repeated root means the equation $s_A - s_B = (t-10)^2$ touches zero at $t=10$ but is positive on both sides of it (since a square is never negative). This means Car A’s displacement is always greater than or equal to Car B’s displacement throughout $t\ge5$ , with equality only at the single instant $t=10$ s. The two cars momentarily draw level, but Car B never actually overtakes Car A (there is no sign change, only a touch).

The distance from $O$ at this instant is found by substituting $t=10$ into either expression:

$s_A(10) = 10^2 = 100$

$s_B(10) = 20(10-5) = 20(5) = 100$

Both give $100$ m, confirming the two cars are $100$ m from $O$ when they draw level.

Final answers

(a) $s_A = \boxed{t^2}$ , $s_B = \boxed{20(t-5)}$ for $t\ge5$
(b) $(t-10)^2=0 \implies t=\boxed{10\text{ s}}$ (a repeated root)
(c) The cars are level at only one instant, $t=10$ s, a distance of $\boxed{100\text{ m}}$ from $O$