A car starts from rest and attains a velocity of 40 metres per second in 20 seconds. it maintains this velocity for 30 seconds and then comes to rest in 25 seconds.
1. sketch the velocity time graph for the motion.
I can't sketch it here.
The first 20 seconds velocity-time will be a sloping graph upward.
the next 30 seconds will be a horizontal line
the last 25 seconds will be a sloping line downward, ending at v=0
Thank you all for your help
a toy car starts from rest and accelerate uniformly at 2.0m/s square for 5.0 seconds. calculate its final velocity...
Please help me with this also
To sketch the velocity-time graph for the given motion, we need to break it down into three phases: acceleration, constant velocity, and deceleration.
Phase 1: Acceleration:
The car starts from rest, meaning its initial velocity is 0 m/s. It then attains a velocity of 40 m/s in 20 seconds. This implies that the car is accelerating uniformly from rest to 40 m/s in 20 seconds. The velocity-time graph during this phase will be a straight line inclined with a positive slope.
Phase 2: Constant Velocity:
The car maintains a velocity of 40 m/s for 30 seconds. This means there is no change in velocity during this phase. Therefore, the velocity-time graph will be a straight horizontal line at 40 m/s.
Phase 3: Deceleration:
The car comes to rest in 25 seconds. This means it decelerates uniformly from 40 m/s to rest (0 m/s) in 25 seconds. The velocity-time graph during this phase will be a straight line inclined with a negative slope.
In summary, the velocity-time graph will have a positive sloping line for acceleration, followed by a flat line for constant velocity, and finally a negative sloping line for deceleration.
Please note that the exact magnitude of the slopes will depend on the scale used for the graph.