the speed of a bus increase uniform from 15mls 6mls in 20s calculate.a the average speed.b accelaration. c the distance travelled during the either period
Increases ?? from 15 m/s to 6 m/s ???
anyway
a = (final velocity - initial velocity) / time
d = Vi t + (1/2) a t^2
= 15 (20) + (1/2) a (400)
Na bro it aint like dat
First subtract the distance (change in velocity)(6-15) , the question is not even correct cause it can't increase from 15 to 6 maybe he or she meant 60 so u subtract (60 from 15) then whatever u get divided by the time which is 20 sec that's how u get the answer for acceleration
To calculate the average speed, acceleration, and distance traveled during the given time period, we need to use the formulas of motion. Let's break down each part of the question and go through the calculations step-by-step.
a) Average speed:
The average speed can be found by taking the total distance traveled and dividing it by the total time taken. In this case, the bus starts at a speed of 15 m/s and increases uniformly to 6 m/s in 20 seconds.
Average speed = (Initial speed + Final speed) / 2
Let's calculate the average speed:
Average speed = (15 m/s + 6 m/s) / 2
Average speed = 21 m/s / 2
Average speed = 10.5 m/s
b) Acceleration:
Acceleration can be found using the formula:
Acceleration = (Change in velocity) / Time
In this case, the change in velocity is the difference between the final speed and initial speed. The time is given as 20 seconds.
Acceleration = (Final speed - Initial speed) / Time
Let's calculate the acceleration:
Acceleration = (6 m/s - 15 m/s) / 20 s
Acceleration = (-9 m/s) / 20 s
Acceleration = -0.45 m/s² (negative because the bus is slowing down)
c) Distance traveled during the entire period:
To find the distance traveled, we can use the formula:
Distance = Initial velocity * Time + 0.5 * Acceleration * Time²
Substituting the given values:
Initial velocity = 15 m/s (initial speed)
Acceleration = -0.45 m/s² (calculated from part b)
Time = 20 seconds
Distance = (15 m/s * 20 s) + (0.5 * -0.45 m/s² * 20 s * 20 s)
Distance = 300 m - 360 m
Distance = -60 m (negative value because of the acceleration being negative)
Therefore, the distance traveled during the given time period is -60 meters, indicating that the bus has moved in the opposite direction.