To draw the velocity-time graph for the given information, we need to break down the motion into different phases and determine the corresponding velocities at different time intervals.
Phase 1: Acceleration for the first 10s
Given: Uniform acceleration of 2 m/s^2 for the first 10s.
Using the equation of motion:
v = u + at
Where:
v = final velocity
u = initial velocity (0 m/s, as the bus starts from rest)
a = acceleration (2 m/s^2)
t = time (10s)
Using the equation, we can calculate the final velocity at the end of this phase:
v = 0 + 2 * 10
v = 20 m/s
So after the first 10s, the velocity of the bus is 20 m/s.
Phase 2: Acceleration for the next 15s
Given: Uniform acceleration of 1 m/s^2 for the next 15s.
Now we need to calculate the final velocity at the end of this phase, considering the initial velocity from the first phase (20 m/s).
Using the same equation as before:
v = u + at
Where:
v = final velocity
u = initial velocity (20 m/s, from the previous phase)
a = acceleration (1 m/s^2)
t = time (15s)
v = 20 + 1 * 15
v = 35 m/s
So after 25s, the velocity of the bus is 35 m/s.
Phase 3: Constant speed for 70s
During this phase, the bus maintains a constant speed. Therefore, the velocity is constant at 35 m/s for the next 70s.
Phase 4: Deceleration for the final 20s
Given: Uniform deceleration to bring the bus to rest in 20s.
For deceleration, we can simply consider the negative sign for acceleration (-1 m/s^2).
Using the equation of motion:
v = u + at
Where:
v = final velocity (0 m/s, as the bus comes to rest)
u = initial velocity (35 m/s, from the previous phase)
a = acceleration (-1 m/s^2)
t = time (20s)
0 = 35 + (-1) * 20
0 = 35 - 20
0 = 15
So after 95s, the velocity of the bus is 0 m/s (i.e., it comes to rest).
Now, we have the velocities at different time intervals:
0 - 10s: 0 m/s (rest)
10 - 25s: 20 m/s (uniform acceleration)
25 - 95s: 35 m/s (constant speed)
95 - 115s: 0 m/s (uniform deceleration)
Using this information, we can plot the velocity-time graph with time on the x-axis and velocity on the y-axis.