Questions: A person travels 75.0 m with a final velocity of 10.0 m / s over 10.0 s. What is the person's acceleration? a=[?] m / s^2

Solution Steps

We are given the following information:

• Final velocity, \( v_f = 10.0 \, \text{m/s} \)
• Time, \( t = 10.0 \, \text{s} \)
• Distance traveled, \( d = 75.0 \, \text{m} \)

We use the kinematic equation: \[ d = v_i t + \frac{1}{2} a t^2 \] where \( v_i \) is the initial velocity and \( a \) is the acceleration. We need to find \( v_i \) first.

We use the equation: \[ v_f = v_i + a t \] Rearranging for \( v_i \): \[ v_i = v_f - a t \]

Substitute \( v_i = v_f - a t \) into the distance equation: \[ d = (v_f - a t) t + \frac{1}{2} a t^2 \] Simplify: \[ d = v_f t - a t^2 + \frac{1}{2} a t^2 \] \[ d = v_f t - \frac{1}{2} a t^2 \]

Rearrange the equation to solve for \( a \): \[ d = v_f t - \frac{1}{2} a t^2 \] \[ 75.0 = 10.0 \times 10.0 - \frac{1}{2} a \times 10.0^2 \] \[ 75.0 = 100.0 - 50.0 a \] \[ 50.0 a = 100.0 - 75.0 \] \[ 50.0 a = 25.0 \] \[ a = \frac{25.0}{50.0} \] \[ a = 0.5000 \, \text{m/s}^2 \]

Final Answer