a car moving at 10ms - was uniformly accelerated for 5s until it velocity reaches 20ms -. it then maintain the speed for another 20s after which the brakes were applied as the vehicle was finally brought to rest in a further 10s.draw the velocity time graph of the motion
v at 10 m/s at t = 0
a = (final velocity-initial velocity)/time
= (20 - 10)/5 = 2 m/s^2
so from (0,10) slope up = 2 m/s^2
to (5,20)
from (5,20) to (25,20) horizontal
from (25,20) to (35, 0 ) straight slope down (slope = -2)
The acceleration, (final velocity-initial velocity)divided by time =10-30 divided by 5=-4
To draw the velocity-time graph, we need to analyze the given information and break it down into three phases:
Phase 1: Uniform Acceleration
The car is initially at rest (0 m/s) and uniformly accelerates for 5 seconds until it reaches a velocity of 20 m/s.
Phase 2: Constant Velocity
The car maintains a constant velocity of 20 m/s for the next 20 seconds.
Phase 3: Braking and Coming to Rest
The brakes are applied, and the car comes to rest (0 m/s) after 10 seconds.
Now, let's plot the graph:
1. Plot Phase 1:
- The initial velocity is 0 m/s at time t = 0.
- The final velocity is 20 m/s after 5 seconds (t = 5).
- During this phase, the velocity increases uniformly, so the graph will be a straight line.
- Connect the points (0,0) and (5,20) with a straight line.
2. Plot Phase 2:
- The velocity remains constant at 20 m/s for the next 20 seconds (t = 5 to t = 25).
- Draw a horizontal line at y = 20 from t = 5 to t = 25.
3. Plot Phase 3:
- At t = 25, the brakes are applied, and the car decelerates until it comes to rest (final velocity of 0 m/s) after 10 seconds (t = 35).
- The graph should smoothly decrease from 20 m/s to 0 m/s over 10 seconds.
- Connect point (25,20) with (35,0) using a smooth curved line.
Overall, the graph will have a positive linear slope during Phase 1, a flat line during Phase 2, and a negative curved slope during Phase 3.