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+5(4)= 20m/s
=0+5(3)= 15m/s
=0+2(3)=6m/s
so I found the veolcity for each of the three parts what would be the final velocity? would it be v-v_0 so 6-20 =-14m/s
b) d=vt
=20*4=80m
=15*3=45m
=6*3=18m
then I add these all up and got 143m as the answer.
No.
The velocity attained in the first period is 20m/s. Then, in the third period, a further deacceleration brings it down 14 m/s.
distance:
The average velocity during the first period is 10m/s, for four seconds: 40m
the average velocity during the second period is 20m/s for 3 seconds, and additional 60 m
The average velocity during the last period is 17 m/s, for three seconds, and additional 17*3 or 51m
a) To find the final speed of the car, you correctly calculated the individual velocities for each part. However, to find the final velocity, you need to consider that the car is decelerating during the third part. When the car decelerates, you need to subtract the acceleration from the initial velocity instead of adding it. So the correct calculation would be:
Final velocity = 6 m/s - 2 m/s^2 (deceleration) * 3 s = 0 m/s
Therefore, the final speed of the car is 0 m/s.
b) To find the distance traveled by the car, you correctly used the equation d = vt for each part and calculated the distance for each section. To find the total distance traveled, you need to add up the distances from each part:
Total distance = 80 m + 45 m + 18 m = 143 m
So the car travels a total distance of 143 meters.
a) To find the final speed of the car, we need to consider the acceleration and deceleration separately.
First, during the acceleration phase of 4 seconds at 5 m/s^2, we can use the equation:
v = v_0 + at
where v_0 is the initial velocity (0 m/s) and a is the acceleration (5 m/s^2). Plugging in the values, we get:
v = 0 + 5(4) = 20 m/s
Next, during the constant velocity phase of 3 seconds, the speed remains the same at 20 m/s.
Finally, during the deceleration phase of 3 seconds at -2 m/s^2 (negative because it's a deceleration), we use the same equation:
v = v_0 + at
where v_0 is the previous velocity (20 m/s) and a is the deceleration (-2 m/s^2). Plugging in the values, we get:
v = 20 - 2(3) = 14 m/s
So, the final speed of the car is 14 m/s.
b) To find the total distance traveled, we can calculate the distance for each phase separately and add them up.
During the acceleration phase, we can use the equation:
d = v_0t + 0.5at^2
where v_0 is the initial velocity (0 m/s), a is the acceleration (5 m/s^2), and t is the time (4 seconds). Plugging in the values, we get:
d = 0(4) + 0.5(5)(4^2) = 40 m
During the constant velocity phase, the distance traveled is simply the speed multiplied by the time:
d = 20 * 3 = 60 m
During the deceleration phase, we can again use the same equation:
d = v_0t + 0.5at^2
where v_0 is the previous velocity (20 m/s), a is the deceleration (-2 m/s^2), and t is the time (3 seconds). Plugging in the values, we get:
d = 20(3) + 0.5(-2)(3^2) = 54 m
Adding up the distances from each phase, we get:
40 + 60 + 54 = 154 m
So, the car travels a total distance of 154 m.