Draw each graph neatly
On its way along a straight line, a car starts from rest and accelerated at a rate of 2.5 m/s² for 10s. It maintained its speed for further 5 s and decelerated by 0.5m/s² to come to rest. What is the time taken for the entire motion? How far did it travel? Draw the position vs time, and velocity vs time graph of the motion of the car
To solve this problem, we can break down the motion of the car into three phases: acceleration, constant velocity, and deceleration.
The given information:
Acceleration (a) = 2.5 m/s²
Time for acceleration (t1) = 10 s
Deceleration (d) = -0.5 m/s² (negative because it's deceleration)
Time for deceleration (t2) = unknown
Phase 1: Acceleration
During acceleration, the car starts from rest, so its initial velocity (v0) is 0 m/s. Using the equation v = v0 + at, we can find the final velocity (v1) at the end of acceleration.
v0 = 0 m/s
a = 2.5 m/s²
t1 = 10 s
v1 = v0 + at1
v1 = 0 + 2.5 * 10
v1 = 25 m/s
Phase 2: Constant Velocity
During this phase, the car maintains its speed for 5 seconds. Since the velocity remains constant, the acceleration is 0 m/s².
Phase 3: Deceleration
At the end of the constant velocity phase, the car decelerates to come to rest. Again using the equation v = v0 + at, we can find the time for deceleration (t2) and the distance traveled during this phase.
v2 = 0 m/s
a = -0.5 m/s²
t2 = unknown
v2 = v1 + a*t2
0 = 25 - 0.5 * t2
0.5 * t2 = 25
t2 = 50 s
To find the total time taken for the entire motion, we add the times for each phase.
Total time = t1 + 5 + t2
Total time = 10 + 5 + 50
Total time = 65 s
To find the distance traveled, we first need to calculate the distance traveled during each phase.
Distance during acceleration = (v1^2 - v0^2) / (2a)
Distance during acceleration = (25^2 - 0^2) / (2 * 2.5)
Distance during acceleration = 625 / 5
Distance during acceleration = 125 m
Distance during constant velocity = v1 * t2
Distance during constant velocity = 25 * 5
Distance during constant velocity = 125 m
Distance during deceleration = (v2^2 - v1^2) / (2d)
Distance during deceleration = (0^2 - 25^2) / (2 * -0.5)
Distance during deceleration = (-625) / (-1)
Distance during deceleration = 625 m
Total distance traveled = Distance during acceleration + Distance during constant velocity + Distance during deceleration
Total distance traveled = 125 + 125 + 625
Total distance traveled = 875 m
Now let's draw the position vs time and velocity vs time graphs:
Position vs Time graph:
^
| *
| *
| *
| *
| *
| *
| *
| *
| *
| *
------------------------>
0 10 15 50 65
Velocity vs Time graph:
^
| ----*----
| / \
| / \
| / \
| -- --
| /
| /
| --
| /
------------------------>
0 10 15 50 65
In the position vs time graph, the car starts from rest and has a steadily increasing position during acceleration, stays at a constant position during constant velocity, and then decreases its position during deceleration until it reaches rest again.
In the velocity vs time graph, the car starts from rest and has a steady increase in velocity during acceleration, maintains a constant velocity during the constant velocity phase, and then has a steady decrease in velocity during deceleration until it reaches rest again.
I hope this helps! Let me know if you have any further questions.