A car going at 30 m/s undergoes acceleration of 2 m/s^2 for 4 sec. What is its final speed? How far did it travel while it was accelerating?
acceleration=the change in velocity/(divided by) the change in time
plz help thx
30m/s at 0 sec
30m/s + 2m/s (at 1 sec)
30m/s + 2m/s + 2m/s (at 2 sec)
etc.
The distance = (1/2)*(acceleration)*(time)^2 + (initial velocity)*(time).
what is the distance?
i got 184 is that correct?
the answer on my worksheet says 136...is this wrong or right?
PLZ HELP!!
136 is correct. Check your work.
To find the final speed of the car, you need to use the equation of motion:
Final velocity (v) = Initial velocity (u) + (acceleration (a) * time (t))
In this case, the initial velocity (u) is 30 m/s, the acceleration (a) is 2 m/s^2, and the time (t) is 4 seconds.
Final velocity (v) = 30 m/s + (2 m/s^2 * 4 s)
Final velocity (v) = 30 m/s + 8 m/s
Final velocity (v) = 38 m/s
Therefore, the final speed of the car is 38 m/s.
Now, to find the distance traveled while accelerating, you can use the equation:
Distance (s) = Initial velocity (u) * time (t) + 0.5 * acceleration (a) * (time (t))^2
In this case, the initial velocity (u) is 30 m/s, the acceleration (a) is 2 m/s^2, and the time (t) is 4 seconds.
Distance (s) = 30 m/s * 4 s + 0.5 * 2 m/s^2 * (4 s)^2
Distance (s) = 120 m + 0.5 * 2 m/s^2 * 16 s^2
Distance (s) = 120 m + 0.5 * 2 m/s^2 * 256 s^2
Distance (s) = 120 m + 2 m/s^2 * 256 s^2
Distance (s) = 120 m + 512 m
Distance (s) = 632 m
Therefore, the car traveled a distance of 632 meters while it was accelerating.