On entering a state highway, a car continues to move with a constant velocity of 6 m s⁻¹ for 2 minutes and then accelerates with a constant acceleration 1 m s⁻² for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.
774 m
Explanation
The textbook context establishes that the area enclosed by the velocity-time graph and the time axis for a desired time interval equals the displacement. The car's motion has two phases. First, constant velocity creates a rectangular area under the graph. Second, constant acceleration creates an area that is the sum of a rectangle and a triangle, exactly as described in the context's example (area of rectangle ACDE + area of triangle ABC).
Solution Steps
Step 1: Convert time to SI units. 2 minutes = s = 120 s. Total time = 126 s.
Step 2: Calculate displacement for the constant velocity phase. Area of rectangle = velocity × time = 6 m s⁻¹ × 120 s = 720 m.
Step 3: Calculate displacement for the constant acceleration phase. Change in velocity = acceleration × time = 1 m s⁻² × 6 s = 6 m s⁻¹. Area = Area of rectangle + Area of triangle = (6 m s⁻¹ × 6 s) + (½ × 6 s × 6 m s⁻¹) = 36 m + 18 m = 54 m.
Step 4: Calculate total displacement. Total Displacement = 720 m + 54 m = 774 m.