A particle moves in a straight line so that its position (x cm) from a fixed point O at time (t seconds) is given by x= t^3-6t^2-15t-5
Find the average speed of the particle in the first 7 seconds
= t^3 - 6t^2 - 15t - 5.
Xo=0^3 - 6*0 - 15*0 - 5 = -5 cm @ t = 0.
X=7^3 - 6*7^2 - 15*7 - 5 = -61 cm @ 7s.
V = D/t = (X-Xo)/t.
V = D/t = (-61-(-5)) / 7 = 56 / 7=-8cm/s
To find the average speed of the particle in the first 7 seconds, we need to calculate the distance traveled by the particle in that time interval and divide it by the time taken.
Given that the position of the particle at time t is given by x = t^3 - 6t^2 - 15t - 5, we can find the distance traveled by taking the difference between the position at the end of the interval and the position at the beginning of the interval.
So, let's calculate the position at the beginning and end of the 7-second interval:
At t = 0 seconds, substitute t = 0 into the equation:
x = (0)^3 - 6(0)^2 - 15(0) - 5
x = 0 - 0 - 0 - 5
x = -5
At t = 7 seconds, substitute t = 7 into the equation:
x = (7)^3 - 6(7)^2 - 15(7) - 5
x = 343 - 294 - 105 - 5
x = -61
Now, calculate the distance traveled by taking the difference between the positions:
Distance = final position - initial position
Distance = -61 - (-5)
Distance = -61 + 5
Distance = -56 cm
The average speed is the distance traveled divided by the time taken:
Average Speed = Distance / Time
Average Speed = -56 cm / 7 seconds
Average Speed = -8 cm/s
Therefore, the average speed of the particle in the first 7 seconds is -8 cm/s.
To find the average speed of the particle in the first 7 seconds, we need to calculate the total distance traveled by the particle in that time period and divide it by the total time.
To find the total distance traveled, we need to integrate the absolute value of the velocity function.
The velocity function is the derivative of the position function with respect to time.
Given that x = t^3 - 6t^2 - 15t - 5, we can find the velocity function by taking the derivative:
v = dx/dt = d/dt(t^3 - 6t^2 - 15t - 5)
Differentiating each term in the equation:
v = 3t^2 - 12t - 15
Now we need to find the absolute value of the velocity function:
|v| = |3t^2 - 12t - 15|
The absolute value of a function takes its positive value regardless of its actual sign. In this case, we are only interested in the magnitude of the velocity.
Next, to find the total distance, we integrate the absolute value of the velocity function over the given time interval (0 to 7 seconds):
distance = ∫(0 to 7) |3t^2 - 12t - 15| dt
Integrating a polynomial function can be done term by term:
distance = ∫(0 to 7) (3t^2 - 12t - 15) dt
distance = ∫(0 to 7) 3t^2 dt - ∫(0 to 7) 12t dt - ∫(0 to 7) 15 dt
Integrating each term:
distance = t^3 - 6t^2 - 15t + C |(0 to 7) - 6(t^2) + C |(0 to 7) - 15(t) + C |(0 to 7)
distance = (7^3 - 6(7)^2 - 15(7)) - (0^3 - 6(0)^2 - 15(0))
distance = (343 - 294 - 105) - (0 - 0 - 0)
distance = 343 - 294 - 105
distance = -56 cm
Since distance cannot be negative, we take the positive value:
distance = 56 cm
Finally, we can find the average speed by dividing the total distance by the total time:
average speed = distance / time
average speed = 56 cm / 7 s
average speed = 8 cm/s
Therefore, the average speed of the particle in the first 7 seconds is 8 cm/s.