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. 20 ft . 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: \[ s(t) = s_0 + v_0 t + \frac{1}{2} a t^2 \] Here, \( s_0 \) is the initial position (0 ft), \( v_0 \) is the initial velocity (1 ft/sec), \( a \) is the constant acceleration (2 ft/sec²), and \( t \) is the time (4 seconds). Plugging in the values: \[ s(4) = 0 + (1)(4) + \frac{1}{2}(2)(4^2) \] \[ s(4) = 4 + \frac{1}{2}(2)(16) = 4 + 16 = 20 \text{ ft} \] Now, to find the velocity at \( t = 4 \) seconds, we can use the velocity equation: \[ v(t) = v_0 + at \] Substituting the known values: \[ v(4) = 1 + (2)(4) = 1 + 8 = 9 \text{ ft/sec} \] So the velocity at \( t = 4 \) seconds is \( 9 \text{ ft/sec} \).