Two cars A and B start moving with a constant acceleration from rest in a straight line. Car A attains a velocity of 5 m s⁻¹ in 5 s. Car B attains a velocity of 3 m s⁻¹ in 10 s. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement mentioned in the two time intervals.
Displacement of Car A = 12.5 m; Displacement of Car B = 15 m
Explanation
The question requires calculating acceleration for both cars, plotting their velocity-time graphs, and then finding displacement as the area under the v-t graph. Since both cars start from rest and move with constant acceleration, their v-t graphs are straight lines passing through the origin. The area under each line (a triangle) gives the displacement for the respective time interval.
Solution Steps
Step 1: Calculate acceleration of Car A. Since it starts from rest (u = 0), acceleration a = (v - u)/t = (5 - 0)/5 = 1 m/s²
Step 2: Calculate acceleration of Car B. Since it starts from rest (u = 0), acceleration a = (v - u)/t = (3 - 0)/10 = 0.3 m/s²
Step 3: Calculate velocities at five instants for Car A (a = 1 m/s²): At t = 0s, v = 0 m/s; t = 1s, v = 1 m/s; t = 2s, v = 2 m/s; t = 3s, v = 3 m/s; t = 4s, v = 4 m/s; t = 5s, v = 5 m/s
Step 4: Calculate velocities at five instants for Car B (a = 0.3 m/s²): At t = 0s, v = 0 m/s; t = 2s, v = 0.6 m/s; t = 4s, v = 1.2 m/s; t = 6s, v = 1.8 m/s; t = 8s, v = 2.4 m/s; t = 10s, v = 3 m/s
Step 5: Plot the v-t graph for both cars on the same axes. Both lines start from the origin. Car A's line is steeper (slope = 1) reaching 5 m/s at 5 s. Car B's line is less steep (slope = 0.3) reaching 3 m/s at 10 s.
Step 6: Calculate displacement of Car A (0 to 5 s). The area under the graph is a triangle: Displacement = (1/2) × base × height = (1/2) × 5 s × 5 m/s = 12.5 m
Step 7: Calculate displacement of Car B (0 to 10 s). The area under the graph is a triangle: Displacement = (1/2) × base × height = (1/2) × 10 s × 3 m/s = 15 m