# Car Motion Problem Solution A car follows this motion: - Travels at \( 20\,\mathrm{m/s} \) for \( 4.\,\mathrm{s} \) (constant speed). Then **accelerates** to \( 30\,\mathrm{m/s} \) over the next \( 4.\,\mathrm{s} \) (constant acceleration). Let's solve each part step-by-step. --- ## a. Acceleration During the 2nd Part ### **Given:** - Initial velocity \( u = 20\,\mathrm{m/s} \) - Final velocity \( v = 30\,\mathrm{m/s} \) - Time \( t = 4.\,\mathrm{s} \) ### **Formula:** \[ a = \frac{v-u}{t} \] ### **Calculation:** \[ a = \frac{30 - 20}{4.} = \frac{10}{4.} = 2.5\,\mathrm{m/s^2} \] ### **Final Answer (a):** \[ \boxed{2.5\,\mathrm{m/s^2}} \] --- ## b. Total Distance Traveled in 8. s There are two intervals: ### **First Interval (constant speed):** - Speed \( v_1 = 20\,\mathrm{m/s} \) - Time \( t_1 = 4.\,\mathrm{s} \) - Distance: \( d_1 = v_1 \times t_1 = 20 \times 4. = 80\,\mathrm{m} \) ### **Second Interval (acceleration):** - Initial speed \( u = 20\,\mathrm{m/s} \) - Acceleration \( a = 2.5\,\mathrm{m/s^2} \) - Time \( t_2 = 4.\,\mathrm{s} \) - Use: \( d_2 = ut_2 + \frac{1}{2}a t_2^2 \) \[ d_2 = 20 \times 4. + \frac{1}{2} \times 2.5 \times (4.)^2 \] \[ d_2 = 80 + .5 \times 2.5 \times 16 \] \[ d_2 = 80 + 1.25 \times 16 \] \[ d_2 = 80 + 20 = 100\,\mathrm{m} \] ### **Total Distance:** \[ d_{total} = d_1 + d_2 = 80 + 100 = 180\,\mathrm{m} \] ### **Final Answer (b):** \[ \boxed{180\,\mathrm{m}} \]