Two cars A and B start moving with a constant acceleration from rest in a straight line. Class 9
Two cars A and B start moving with a constant acceleration from rest in a straight line. Class 9
Question 1.
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. (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph). Class 9
Answer:
Step 1: Find accelerations For Car A:
a_A = (v - u) /1
= (5 - 0)/5
= 1 m s-2
For Car B:
a_B = (v - u) /1
= (3 - 0)/10
= 0.3 m s-2
Step 2: Calculate velocities at five instants of time for each car
| Time (s) | Velocity of A (m s-1) v = 1 × t | Velocity of B (m s-1) v = 0.3 × t |
| 0 | 0 | 0 |
| 2 | 2 | 0.6 |
| 4 | 4 | 1.2 |
| 6 | 6 | 1.8 |
| 8 | 8 | 2.4 |
| 10 | 10 | 3.0 |
Velocity-time graph description:
- Both lines start at origin (0, 0)
- Car A: steeper straight line (reaches 5 m s-1 at 5 s, 10 m s-1 at 10 s)
- Car B: less steep straight line (reaches 3 m s-1 at 10 s)
Step 3: Calculate displacement
For Car A in 5 s (time interval of Car A reaching 5 m s-1):
s_A = 1/2 × (u + v) × t
= 1/2 × (0 + 5) × 5
= 12.5 m
For Car B in 10 s (time interval of Car B reaching 3 m s-1):
s_B = 1/2 × (u + v) × t
= 1/2 × (0 + 3) × 10
= 15 m
(These values equal the areas of the triangles under each line in the velocity-time graph.)