A remote control car is traveling in a straight line with a constant acceleration of \( 2 \mathrm{ft} / \mathrm{sec}^{2} \). If the initial position (at \( t=0 \) ) is 0 ft and the initial velocity is \( 1 \mathrm{ft} / \mathrm{sec} \), find the position at \( t=4 \) seconds. [ ? ] ft .

To find the position of the remote control car at \( t = 4 \) seconds, we can use the kinematic equation for position: \[ s(t) = s_0 + v_0 t + \frac{1}{2} a t^2 \] where: - \( s(t) \) is the position at time \( t \), - \( s_0 \) is the initial position, - \( v_0 \) is the initial velocity, - \( a \) is the acceleration. Given: - \( s_0 = 0 \) ft (initial position), - \( v_0 = 1 \) ft/s (initial velocity), - \( a = 2 \) ft/s² (acceleration), - \( t = 4 \) seconds. Now we can plug in the values: \[ s(4) = 0 + (1)(4) + \frac{1}{2}(2)(4^2) \] Calculating \( s(4) \): \[ s(4) = 4 + \frac{1}{2}(2)(16) \] \[ s(4) = 4 + (1)(16) \] \[ s(4) = 4 + 16 \] \[ s(4) = 20 \text{ ft} \] Thus, the position of the remote control car at \( t = 4 \) seconds is \( 20 \) ft.