A car starts from rest and accelerates at 5 m/s^2 for 4 s, then maintains that velocity for 3 seconds, then decelerates at the rate of 2m/s^2 for 3 seconds.
a) what is the final speed of the car?
b) how far does the car travel?
a) v = v_0 +at
=0+2(3)= 6m/s
Would I use the part of the question where the car decelerates to 2 m/s for 3 s? Or would t=10s and a=2m/s^s?
b) d= 1/2at^2
=1/2 (6)(10)^2
=1/2(600)
=300m
I got 6m/s^2 as the acceleration from question a)
Is this correct?
a) find the first acceleration:
vf=vo+at= 5*4= 20m/s, then it slows
vf=20m/s-2*3= 14m/s
how far: break it up into three parts again, and solve.
a) To find the final speed of the car, we need to consider the three different segments of motion separately. We start from rest and accelerate at 5 m/s^2 for 4 seconds. During this segment, the initial velocity is 0 m/s and the acceleration is 5 m/s^2. We can use the equation v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time.
Using this equation, the final velocity during the acceleration phase is v = 0 + 5(4) = 20 m/s.
Next, the car maintains that velocity for 3 seconds. Since the velocity is constant during this time, the final velocity is still 20 m/s.
Finally, the car decelerates at a rate of 2 m/s^2 for 3 seconds. Here, the initial velocity is 20 m/s and the acceleration is -2 m/s^2 (negative because it's decelerating). Using the same equation as before, we have v = 20 + (-2)(3) = 14 m/s.
Therefore, the final speed of the car is 14 m/s.
b) To find the distance traveled by the car, we can sum up the distances covered during each segment of motion. In the first segment, the car is accelerating, so we can use the equation d = 1/2at^2, where d is the distance, a is the acceleration, and t is the time.
Using this equation, the distance covered during the acceleration phase is 1/2(5)(4)^2 = 40 m.
During the constant velocity phase, the car travels at a speed of 20 m/s for 3 seconds. So the distance covered here is 20 m/s * 3 s = 60 m.
During deceleration, we can again use the equation d = 1/2at^2, but this time the acceleration is -2 m/s^2 and the time is 3 seconds. Thus, the distance covered during deceleration is 1/2(-2)(3)^2 = -9 m.
Note that the distance during deceleration is negative because the car is moving backward. This negative distance represents the 'undoing' of the distance that the car covered during acceleration.
To find the total distance, we add up the distances covered during each segment: 40 m + 60 m + (-9) m = 91 m.
Therefore, the car travels a total distance of 91 meters.
In conclusion:
a) The final speed of the car is 14 m/s.
b) The car travels a total distance of 91 meters.