# Sprinter Race Problem **Given:** - Total distance (\(d\)) = 100 m - Total time (\(T\)) = 8.5 s - Initial velocity (\(u\)) = - Top speed (\(v\)) = 15 m/s - Time to reach top speed = \(t\) - Acceleration during \(t\): \(a\) (constant) - After reaching 15 m/s, he continues at that speed. --- ## a) How long did it take for the sprinter to reach his top speed? (\(t\)) ### 1. **Distance covered during acceleration (\(d_1\)):** From rest, with constant acceleration, \[ d_1 = \frac{1}{2} a t^2 \] But since final speed after \(t\) seconds is 15 m/s, \[ v = u + a t \implies 15 = + a t \implies a = \frac{15}{t} \] So, \[ d_1 = \frac{1}{2} a t^2 = \frac{1}{2} \left(\frac{15}{t}\right) t^2 = \frac{15}{2} t \] ### 2. **Distance covered at constant speed (\(d_2\)):** After accelerating for \(t\) seconds, he runs at 15 m/s for the rest of the race. Time at constant speed: \[ \text{Time at constant speed} = T - t \] Distance: \[ d_2 = v \cdot (T - t) = 15 (8.5 - t) \] ### 3. **Total distance equation:** \[ d = d_1 + d_2 \] \[ 100 = \frac{15}{2} t + 15(8.5 - t) \] ### 4. **Solve for \(t\):** \[ 100 = \frac{15}{2} t + 127.5 - 15t \] \[ 100 = 127.5 - 15t + \frac{15}{2} t \] \[ 100 = 127.5 + \left(\frac{15}{2} - 15\right)t \] \[ 100 = 127.5 - \left(\frac{15}{2}\right)t \] \[ \left(\frac{15}{2}\right)t = 127.5 - 100 = 27.5 \] \[ t = \frac{27.5 \times 2}{15} = \frac{55}{15} = 3.666... \] \[ \boxed{t = 3.67~\text{s}~(\text{to 2 decimal places})} \] --- ## b) What was his acceleration from rest until he reached his top speed? (\(a\)) Recall: \[ a = \frac{15}{t} \] Plug in the value from (a): \[ a = \frac{15}{3.666...} \approx 4.09~\text{m/s}^2 \] \[ \boxed{a = 4.09~\text{m/s}^2~(\text{to 2 decimal places})} \] --- ## **Summary Table** | Part | Answer | |------|-----------------------------| | a) | \( t = 3.67~\text{s} \) | | b) | \( a = 4.09~\text{m/s}^2 \) |

# Sprinter Race Problem **Given:** - Total distance (\(d\)) = 100 m - Total time (\(T\)) = 8.5 s - Initial velocity (\(u\)) = 0 m/s - Top speed (\(v\)) = 15 m/s - Time to reach top speed = \(t\) - Acceleration during \(t\): \(a\) (constant) --- ## a) How long did it take for the sprinter to reach his top speed? (\(t\)) ### 1. **Distance covered during acceleration (\(d_1\))** From rest, with constant acceleration: \[ d_1 = \frac{1}{2} a t^2 \] Final speed after \(t\) seconds: \[ v = u + a t \implies 15 = 0 + a t \implies a = \frac{15}{t} \] Substituting \(a\) into the distance formula: \[ d_1 = \frac{1}{2} \left(\frac{15}{t}\right) t^2 = \frac{15}{2} t \] ### 2. **Distance covered at constant speed (\(d_2\))** After accelerating for \(t\) seconds, he runs at 15 m/s for the remainder of the race: \[ \text{Time at constant speed} = T - t = 8.5 - t \] Distance during this time: \[ d_2 = v \cdot (T - t) = 15 (8.5 - t) \] ### 3. **Total distance equation** Combining both distances: \[ d = d_1 + d_2 \] \[ 100 = \frac{15}{2} t + 15(8.5 - t) \] ### 4. **Solve for \(t\)** Rearranging the equation: \[ 100 = \frac{15}{2} t + 127.5 - 15t \] \[ 100 = 127.5 - 15t + \frac{15}{2} t \] Combine like terms: \[ 100 = 127.5 - \left(15 - \frac{15}{2}\right)t \] \[ 100 = 127.5 - \frac{15}{2}t \] Rearranging gives: \[ \frac{15}{2}t = 127.5 - 100 = 27.5 \] Solving for \(t\): \[ t = \frac{27.5 \times 2}{15} = \frac{55}{15} \approx 3.67 \] \[ \boxed{t = 3.67~\text{s}~(\text{to 2 decimal places})} \] --- ## b) What was his acceleration from rest until he reached his top speed? (\(a\)) Using the formula for acceleration: \[ a = \frac{15}{t} \] Substituting the value of \(t\) from part (a): \[ a = \frac{15}{3.67} \approx 4.09~\text{m/s}^2 \] \[ \boxed{a = 4.09~\text{m/s}^2~(\text{to 2 decimal places})} \] --- ## **Summary Table** | Part | Answer | |------|-----------------------------| | a) | \( t = 3.67~\text{s} \) | | b) | \( a = 4.09~\text{m/s}^2 \) |