A car takes up from west and accelerate uniformly at the rate of 4m/s square until it reaches a velocity of 10m/s. It then maintains this speed for 5secs and later decelerate at the rate of 2m/s square till it reaches its destination
a) draw the velocity time graph of the motion
b) compute the time taken for the car to reach the velocity of 10m/s and the time taken decelerate
c) find the total distance covered in the course of this motion.
time to accelerate
10m/s / 4m/s^2 = 2.5s
deceleration takes twice as long, since the rate is half as great.
Now figure the distance using the good old formula
s = vot + 1/2 at^2
for each part of the trip.
a) To draw the velocity-time graph, we need to understand the different phases of motion.
1. Acceleration phase: The car accelerates uniformly from rest until it reaches a velocity of 10 m/s.
2. Constant velocity phase: The car maintains a constant velocity of 10 m/s for 5 seconds.
3. Deceleration phase: The car decelerates uniformly until it reaches rest at its destination.
Let's break down the graph into these three sections:
1. Acceleration phase:
During the acceleration phase, the car starts from rest and accelerates uniformly at 4 m/s² until it reaches a velocity of 10 m/s. The graph will be a straight line with a positive slope.
2. Constant velocity phase:
During this phase, the car maintains a constant velocity of 10 m/s for 5 seconds. The graph will be a straight horizontal line at the height of 10 m/s.
3. Deceleration phase:
During the deceleration phase, the car decelerates uniformly at 2 m/s² until it comes to rest at its destination. The graph will be a straight line with a negative slope.
b) To compute the time taken for the car to reach a velocity of 10 m/s and the time taken to decelerate, we can use the basic kinematic equation:
v = u + at
where:
v = final velocity
u = initial velocity (in this case, 0 m/s)
a = acceleration
t = time
1. Time taken to reach 10 m/s:
Using the equation, we can plug in the values:
10 = 0 + 4t
t = 2.5 seconds
2. Time taken to decelerate to rest:
Using the same equation, but now with the deceleration (-2 m/s²) and the final velocity (0 m/s):
0 = 10 + (-2)t
t = 5 seconds
c) To find the total distance covered in the course of this motion, we need to calculate the distance traveled during each phase and add them together.
1. Distance during acceleration phase:
Using the kinematic equation:
s = ut + 0.5at²
where:
s = distance
u = initial velocity (in this case, 0 m/s)
a = acceleration
t = time
During the acceleration phase, we know the acceleration is 4 m/s² and the time taken is 2.5 seconds. Plugging these values into the equation:
s = 0 * 2.5 + 0.5 * 4 * (2.5)²
s = 0 + 0.5 * 4 * 6.25
s = 0 + 12.5
s = 12.5 meters
2. Distance during constant velocity phase:
During this phase, the car maintains a constant velocity of 10 m/s for 5 seconds. Therefore, the distance covered is:
s = v * t
s = 10 * 5
s = 50 meters
3. Distance during deceleration phase:
Using the same equation as in the acceleration phase, but now with the deceleration (-2 m/s²) and the time taken (5 seconds):
s = 10 * 5 + 0.5 * (-2) * (5)²
s = 50 + 0.5 * (-2) * 25
s = 50 + (-25)
s = 25 meters
Adding the distances from each phase:
Total distance = Distance during acceleration + Distance during constant velocity + Distance during deceleration
Total distance = 12.5 + 50 + 25
Total distance = 87.5 meters
Therefore, the total distance covered in the course of this motion is 87.5 meters.