A car travels along a straight road for 30 seconds starting at time t = 0. Its acceleration in ft/sec2 is given by the linear graph below for the time interval [0, 30]. At t = 0, the velocity of the car is 0 and its position is 10.
What is the total distance the car travels in this 30 second interval? Your must show your work but you may use your calculator to evaluate. Give 3 decimal places in your answer and include units.
To find the total distance traveled by the car, we need to calculate the area under the graph of acceleration (which represents the change in velocity) as a function of time.
Since the acceleration is given by a linear graph, it means that the rate of change of velocity is constant. In this case, we can use the equation of motion to calculate the change in velocity:
Δv = a * Δt
where Δv is the change in velocity, a is the acceleration, and Δt is the time interval.
In this problem, the acceleration graph is a straight line, which means it is a constant. Let's call it a1 and calculate it using the given information:
Given:
Initial velocity (at t = 0): v_i = 0 ft/s
Position (at t = 0): x_i = 10 ft
Since the velocity is the derivative of position, we know that the initial velocity is the rate of change of position at t = 0. That means:
v_i = dx/dt
Since v_i = 0, we can write:
dx = 0 * dt
dx = 0 ft
This means that the car is not moving at t = 0, so the position doesn't change.
Now, let's calculate the acceleration at each point on the graph:
Acceleration at t = 0: a_0 = 20 ft/s^2
Acceleration at t = 30: a_30 = -10 ft/s^2
To calculate the change in velocity during the 30-second interval, we use the equation:
Δv = a_avg * Δt
where a_avg is the average acceleration during the interval. To find the average acceleration, we sum up the initial and final accelerations and divide by 2:
a_avg = (a_0 + a_30) / 2
a_avg = (20 + (-10)) / 2
a_avg = 5 ft/s^2
Now, substitute the values into the equation:
Δv = a_avg * Δt
Δv = 5 ft/s^2 * 30 s
Δv = 150 ft/s
The change in velocity is 150 ft/s.
To find the final velocity, we add the change in velocity to the initial velocity:
v_f = v_i + Δv
v_f = 0 + 150 ft/s
v_f = 150 ft/s
To find the total distance traveled, we need to calculate the area under the graph of velocity (which represents the change in position) as a function of time.
Since the velocity is a straight line, the area under the graph is a trapezoid. The formula to calculate the area of a trapezoid is:
A = (b1 + b2) * h / 2
where b1 and b2 are the lengths of the parallel sides of the trapezoid, and h is the height.
In this case, the parallel sides are the initial and final velocities, and the height is the time interval.
Substituting the values into the formula:
A = (v_i + v_f) * Δt / 2
A = (0 + 150 ft/s) * 30 s / 2
A = (150 ft/s) * 30 s / 2
A = 2250 ft
The area under the graph, which represents the total distance traveled by the car, is 2250 ft.
To find the total distance traveled by the car in the 30-second interval, we need to find the area under the acceleration-time graph.
Since the acceleration is given by a linear graph, we can represent it using the equation:
a(t) = mt + b
where m is the slope of the graph and b is the y-intercept.
From the graph, the acceleration at t = 0 is 2 ft/sec^2, and at t = 30, it is 8 ft/sec^2.
Using the given information, we can find the values of m and b:
a(0) = 2 ft/sec^2 -> 2 = 0 * m + b -> b = 2
a(30) = 8 ft/sec^2 -> 8 = 30 * m + 2 -> 30m = 6 -> m = 0.2 ft/sec^2
Now we have the equation for acceleration:
a(t) = 0.2t + 2
To find the velocity, we need to integrate the acceleration equation with respect to time:
v(t) = ∫a(t) dt (from t = 0 to t = 30)
v(t) = ∫(0.2t + 2) dt
v(t) = 0.1t^2 + 2t + C
At t = 0, the velocity of the car is 0, so we can find the value of C:
0 = 0.1(0)^2 + 2(0) + C -> C = 0
The equation for velocity becomes:
v(t) = 0.1t^2 + 2t
To find the position, we need to integrate the velocity equation with respect to time:
s(t) = ∫v(t) dt (from t = 0 to t = 30)
s(t) = ∫(0.1t^2 + 2t) dt
s(t) = 0.0333t^3 + t^2 + D
At t = 0, the position of the car is 10 ft, so we can find the value of D:
10 = 0.0333(0)^3 + (0)^2 + D -> D = 10
The equation for position becomes:
s(t) = 0.0333t^3 + t^2 + 10
To find the total distance traveled, we need to calculate the change in position from t = 0 to t = 30:
Distance traveled = s(30) - s(0)
= (0.0333(30)^3 + (30)^2 + 10) - (0.0333(0)^3 + (0)^2 + 10)
= (0.0333(27000) + 900 + 10) - 10
= (897 + 900 + 10) - 10
= 1807 feet (rounded to 3 decimal places)
Therefore, the total distance the car travels in the 30-second interval is 1807 feet.