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'll break the motion of the car into three phases: acceleration, constant velocity, and deceleration.
Phase 1: Acceleration
The car starts from rest and accelerates at a rate of 2.5 m/s² for 10 seconds. We can use the kinematic equation to find the final velocity (v) during this phase:
v = u + at
v = 0 + 2.5(10)
v = 25 m/s
So, the car's velocity at the end of this phase is 25 m/s.
Phase 2: Constant velocity
The car maintains its speed for another 5 seconds. During this phase, the velocity remains constant at 25 m/s.
Phase 3: Deceleration
The car then decelerates at a rate of 0.5 m/s² until it comes to rest (v = 0 m/s). We can use the kinematic equation again, this time to find the time (t) taken for the car to come to rest:
v = u + at
0 = 25 - 0.5t
0.5t = 25
t = 50 seconds.
So, the car takes 50 seconds to come to rest during this phase.
Now, let's find the total time taken for the entire motion:
Total time = time for acceleration phase + time for constant velocity phase + time for deceleration phase
Total time = 10 + 5 + 50
Total time = 65 seconds
Therefore, the car takes 65 seconds for the entire motion.
To find the distance traveled, we need to calculate the distance covered in each phase.
Phase 1: Acceleration
To find the distance covered during the acceleration phase, we use the equation s = ut + 0.5at²:
s = 0(10) + 0.5(2.5)(10)²
s = 0 + 0.5(2.5)(100)
s = 0 + 0.5(250)
s = 125 m
Phase 2: Constant velocity
During this phase, the car maintains its speed of 25 m/s for 5 seconds:
s = v*t
s = 25*5
s = 125 m
Phase 3: Deceleration
To find the distance covered during the deceleration phase, we can use the equation v² = u² + 2as, where v = 0 m/s and u = 25 m/s:
0 = 25² + 2*(-0.5)*s
0 = 625 - s
s = 625 m
Total distance covered = distance during acceleration phase + distance during constant velocity phase + distance during deceleration phase
Total distance covered = 125 + 125 + 625
Total distance covered = 875 m
Therefore, the car travels a distance of 875 meters in total.
Now, let's draw the position vs time graph and velocity vs time graph for the motion of the car.
Position vs Time Graph:
During the acceleration phase, the position increases rapidly. During the constant velocity phase, the position remains constant. Finally, during the deceleration phase, the position decreases rapidly until it reaches zero. The graph would look like this:
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0 10 15 65 (time in seconds)
Velocity vs Time Graph:
During the acceleration phase, the velocity increases linearly. During the constant velocity phase, the velocity remains constant at 25 m/s. Finally, during the deceleration phase, the velocity decreases linearly until it reaches zero. The graph would look like this:
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0 10 15 65 (time in seconds)