Question

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.

Answer 1

Answer:

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

Answer 2

Answer:

isn't this physics? im sorry if it's not