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 \[ t=4 \text { seconds. } 20 \mathrm{ft} \text {. } \] What is the velocity at \( t=4 \) seconds? [? ] \( \mathrm{ft} / \mathrm{sec} \)

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 (0 ft), - \( v_0 \) is the initial velocity (1 ft/s), - \( a \) is the acceleration (2 ft/s²), - \( t \) is the time (4 seconds). Plugging in the values: \[ s(4) = 0 + (1)(4) + \frac{1}{2}(2)(4^2) = 4 + \frac{1}{2}(2)(16) = 4 + 16 = 20 \, \text{ft} \] So, the position at \( t = 4 \) seconds is \( 20 \) ft. Now to find the velocity at \( t = 4 \) seconds, we can use the formula for velocity: \[ v(t) = v_0 + a t \] Plugging in the known values: \[ v(4) = 1 + 2(4) = 1 + 8 = 9 \, \text{ft/s} \] So, the velocity at \( t = 4 \) seconds is \( 9 \, \text{ft/s} \).