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 velocity of the car when t = 6? You must show your work and include units in your answer.

To find the velocity of the car at t = 6, we need to integrate the acceleration graph over the interval [0,6] to get the change in velocity.

Given that the acceleration is linear, we can find the equation of the line represented by the graph. We know that the initial velocity at t = 0 is 0, and the final velocity at t = 30 is 7 ft/sec, according to the acceleration graph.

Using this information, we can calculate the slope of the line, which represents the acceleration. The formula for the slope of a line is given by:

slope = (change in y) / (change in x)

In this case, the change in y is 7 ft/sec (final velocity at t = 30) and the change in x is 30 sec (the time interval). Therefore:

slope = (7 ft/sec) / (30 sec) = 7/30 ft/sec^2

Now that we have the acceleration, we can integrate it to find the change in velocity over the interval [0,6]. The formula for integrating acceleration to find velocity is:

∫(acceleration) dt = ∫(7/30) dt

To integrate, we multiply the acceleration by time:

= (7/30) ∫dt from 0 to 6

= (7/30) * [t] from 0 to 6

= (7/30) * (6 - 0)

= (7/30) * 6

= 42/30 ft/sec

Simplifying the fraction, we get:

= 7/5 ft/sec

Therefore, the velocity of the car at t = 6 is 7/5 ft/sec.

To find the velocity of the car when t = 6, we need to integrate the acceleration function with respect to time to obtain the velocity function.

Given that the acceleration function is linear, we can find the equation of the line using the two given points: (0, 0) and (30, a), where a is the acceleration at t = 30.

Using the equation of a line in slope-intercept form, y = mx + b, where m represents the slope and b represents the y-intercept, we have:
m = (a - 0) / (30 - 0) = a / 30

Using the point-slope form, substituting t = 0 and a = 6:
y - 0 = (a / 30) * (t - 0)
y = (a / 30) * t

Since the initial velocity is 0 when t = 0 and the initial position is 10 at t = 0, we can write the velocity function as:
v(t) = (a / 30) * t + 0

Now, we need to find the constant a, which represents the acceleration at t = 30.

Given that the position function is the integral of the velocity function, we can integrate the velocity function over the interval [0, 30] to obtain the position function.

Using the fundamental theorem of calculus, we have:
s(t) = ∫[0,30] v(t) dt
s(t) = ∫[0,30] (a / 30) * t dt
s(t) = (a / 30) * ∫[0,30] t dt
s(t) = (a / 30) * (t^2 / 2) | [0,30]
s(t) = (a / 30) * (30^2 / 2) - (a / 30) * (0^2 / 2)
s(t) = (a / 30) * (900 / 2)
s(t) = (a / 30) * 450
s(t) = 15a

Given that s(30) = 10, we can substitute t = 30 and s(t) = 10 to find a:
10 = 15a
a = 10 / 15
a = 2/3 ft/sec^2

Now that we have the value of a, we can substitute it into the velocity function at t = 6 to find the velocity of the car when t = 6:

v(t) = (2/3 / 30) * t + 0
v(6) = (2/3 / 30) * 6 + 0
v(6) = (2/3 / 30) * 6
v(6) = (2/3) * (1/5)
v(6) = 2/15 ft/sec

Therefore, the velocity of the car when t = 6 is 2/15 ft/sec.

assuming the graph of acceleration is the line

a = 6mt
so,
v = 3mt^2 + c
v(0)=0, so c=0
x = mt^3 + c
x(0) = 10, so c=10, and

x(t) = mt^3 + 10

so, once you decide on m, you can figure x(6)