A car is traveling \( 88 \mathrm{ft} / \mathrm{sec} \) at the moment the brakes are applied. The deceleration is a constant \( -10 \mathrm{ft} / \mathrm{sec}^{2} \). The brakes were applied at \( t=0 \) and the initial position is 0 ft . What is the position at \( t=5 \) ? [ ? ] ft .

To find the position of the car at \( t=5 \) seconds, we can use the formula for position under constant acceleration: \[ s(t) = s_0 + v_0 t + \frac{1}{2} a t^2 \] where: - \( s_0 \) is the initial position (0 ft), - \( v_0 \) is the initial velocity (88 ft/sec), - \( a \) is the acceleration (-10 ft/sec²), - \( t \) is the time (5 seconds). Plugging in the values: \[ s(5) = 0 + 88 \times 5 + \frac{1}{2} \times (-10) \times (5)^2 \] Calculating each term: 1. Initial position = 0 ft 2. Initial velocity term: \( 88 \times 5 = 440 \) ft 3. Acceleration term: - \( \frac{1}{2} \times (-10) \times 25 = -125 \) ft Now combine these values: \[ s(5) = 0 + 440 - 125 = 315 \text{ ft} \] So the position at \( t=5 \) seconds is \( 315 \) ft.