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.

Im not really sure how to go about this? Would I use the trapezoidal rule i dont know please help.

The total distance the car travels in this 30 second interval is 1333.34 units

From the graph, we have the following points

(0, 10) and (10, 0).

Start by calculating the slope (m) of the graph

So, we have:

The equation is then calculated as:

This gives

The above equation represents the acceleration (y) as a function of time (x).

Integrate to get the velocity (v)

From the question, we have:

The velocity (v) of the car is 0, when the time (x) is 0.

So, we have:

This gives

So, the equation becomes

Set v = 0.

So, we have:

Multiply through by -2

Factorize

Split

Solve for x

or  

Integrate velocity (v) to get the displacement (d)

From the question, we have:

The position (d) of the car is 10, when the time (x) is 0.

So, we have:

So, the equation becomes

The position at 30 seconds is:

The position at 20 seconds is:

The total distance is then calculated as:

This gives

Hence, the total distance is 1333.34 units

Read more about distance at:

link

666.667 feet

Step-by-step explanation:

Slope = -1

Intercept = 10

y = -t + 10

y is the acceleration

Integrate y fornv

v = -t²/2 + 10t + c

At t=0, v=0 so c = 0

v = -t²/2 + 10t

Turns when v = 0,

-t²/2 + 10t = 0

t = 0, 20

Integrate v for s

s = -t³/6 + 5t² + c

At t = 0, s = 10

10 = c

s = -t³/6 + 5t² + 10

s at t=30,

-(30³)/6 + 5(30)² + 10

= 10m

(Back to starting point)

At t = 20,

Displacement in

-(20³)/6 + 5(20)² + 10

= 343.333

Total distance = 2(343.333-10)

= 666.6667