A car accelerat uniformly 20m/s to 50m/s in 6 sec what is the distance travelde by the car during the 5th sec ?
a = (50-20)/6
s(t) = 20t + 1/2 at^2
So, you want s(5)-s(4)
Hellp
To calculate the distance traveled by the car during the 5th second, we need to find the average velocity during that time and then multiply it by the time interval.
Given:
Initial velocity (u) = 20 m/s
Final velocity (v) = 50 m/s
Time taken (t) = 6 seconds
First, we can calculate the acceleration (a) using the formula:
a = (v - u) / t
Plugging in the values, we have:
a = (50 - 20) / 6
a = 30 / 6
a = 5 m/s²
Now, since the car is accelerating uniformly, the average velocity during the 5th second (v_avg) can be calculated using the formula:
v_avg = u + (a * t_avg)
where t_avg is the average time during the 5th second. To find t_avg:
t_avg = (t_start + t_end) / 2
where t_start is the start time of the 5th second and t_end is the end time of the 5th second.
In this case, the start time of the 5th second is 4 seconds (since the time intervals start from 0).
t_avg = (4 + 5) / 2
t_avg = 9 / 2
t_avg = 4.5 seconds
Now, we can calculate v_avg:
v_avg = 20 + (5 * 4.5)
v_avg = 20 + 22.5
v_avg = 42.5 m/s
Finally, to determine the distance traveled during the 5th second (s_5th_sec), we use the formula:
s_5th_sec = u * t_avg + (1/2) * a * t_avg²
Plugging in the values we have:
s_5th_sec = 20 * 4.5 + (1/2) * 5 * (4.5)²
s_5th_sec = 90 + (1/2) * 5 * 20.25
s_5th_sec = 90 + (1/2) * 101.25
s_5th_sec = 90 + 50.625
s_5th_sec = 140.625 meters
Therefore, the car has traveled approximately 140.625 meters during the 5th second.