A car starts from rest and accelerates uniformly to 20 m s⁻¹ in 5 seconds. It then travels at 20 m s⁻¹ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.
310 m
Explanation
The problem involves calculating the distance covered in three phases of motion: acceleration, constant velocity, and deceleration. Using the kinematic equation s = (u+v)/2 * t for the accelerating and decelerating phases, and s = v*t for the constant velocity phase, we find the individual distances to be 50 m, 200 m, and 60 m respectively. Their sum gives the total distance.
Solution Steps
Step 1: Calculate distance during acceleration: s₁ = (0 + 20 m s⁻¹) / s = 50 m
Step 2: Calculate distance during constant velocity: s₂ = 20 m s⁻¹ × 10 s = 200 m
Step 3: Calculate distance during deceleration: s₃ = (20 m s⁻¹ + 0 m s⁻¹) / s = 60 m
Step 4: Add all distances: Total distance = 50 m + 200 m + 60 m = 310 m