A bus moves from rest with a uniform acceleration of 2m/s(square) for the first 10s. It then accelerate at a uniform rate of 1m/s(square) for another 15s, it continues at constant speed for 70s and finally comes to rest in 20s by uniform deceleration.Draw the velocity time graph using the information above.
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answer
V=u+at
=0+2×10
=20ms-¹
V1= 20+1×15=35ms-¹
So the total distance traveled
Area of OAG =½×10×20=100m
Area of trapezium GABF
=½(20+30)×15=412.5m
Area of FBCE =35×70=2450m
Area of triangle ECD
=½×20×35=350m
Total distance = 100+412.5+2450+350=3312.5m
Ifedinna
Why did the bus start exercising? Because it wanted to get in shape! Alright, let's draw that velocity-time graph for our hardworking bus.
During the first 10 seconds, the bus accelerates at a uniform rate of 2m/s². So, the velocity increases by 2m/s every second.
In the next 15 seconds, the bus accelerates at a rate of 1m/s². So, the velocity increases by 1m/s every second.
For the next 70 seconds, the bus moves at a constant speed, which means the velocity remains constant.
Finally, the bus comes to a rest in 20 seconds by uniform deceleration. So, the velocity decreases by the same amount every second.
Let's put it all together:
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And there you have it! The velocity-time graph for our productive bus. Keep up the great work, bus!
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.
3312.5
v(0)=0
v(10)=1/2 a t^2=1/2 *2*100=100m/s
so draw a straight line from 0,0 to 10,100
v(15)=100+1/2 a (t-10)^2 =100+ 1/2 (1)(25=125m/s
so plot the point (15,125)
then keep speed a constant horizontal line until time=70+15
plot the point (105, 125)
finally velocity at time125, v=0. Plot that point.