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 position of the car when t = 20? You must show your work answer to 3 decimal places. Include units in your answer.
I'm sure this problem isn't all that difficult, but I'm having a lot of trouble solving. Please help!
I cannot see the graph, but you have the graph for a(t)
v(t) = ∫ a(t) dt + C
But, since v(0) = 0, C=0
Now you know that the position
s(t) = ∫ v(t) dt + C
Since s(0) = 10, you can solve that for C.
Now just find s(20)
I'm still confused, how exactly am I supposed to plug this in?
sigh. what does the graph of the acceleration look like?
Is it constant? a step function? a sloping line?
piecewise? Give me the equation of the line(s) so I can know what to do.
Unless you give me something to go on, there's not much I can do.
review your section on antiderivatives. That will explain how to go from acceleration to velocity, and thence to position.
I apologize, but the website isn't letting me put in the graph. It has a -1 slope and is a decreasing linear graph. Everything else is given in the problem.
To find the position of the car when t = 20, we can use the kinematic equation for position with constant acceleration:
x = x₀ + v₀t + 0.5at²
Where:
x is the position of the car at time t
x₀ is the initial position of the car (given as 10 ft)
v₀ is the initial velocity of the car (given as 0 ft/sec)
t is the time in seconds
a is the acceleration of the car in ft/sec²
First, let's find the acceleration of the car at t = 20. From the given linear graph, we can see that the acceleration is constant. Let's find the slope of the graph to determine the acceleration.
Acceleration = Δv / Δt
Looking at the graph, we can see that the velocity changes from 20 ft/sec to 40 ft/sec in a time interval of 10 seconds (from t = 10 to t = 20). Therefore:
Acceleration = (40 ft/sec - 20 ft/sec) / (10 sec - 0 sec)
Acceleration = 20 ft/sec / 10 sec
Acceleration = 2 ft/sec²
Now that we have the acceleration, we can use the kinematic equation to find the position of the car at t = 20:
x = x₀ + v₀t + 0.5at²
Plugging in the known values:
x = 10 ft + 0 ft/sec * 20 sec + 0.5 * 2 ft/sec² * (20 sec)²
x = 10 ft + 0 ft + 0.5 * 2 ft/sec² * 400 sec²
x = 10 ft + 0 ft + 0.5 * 2 ft/sec² * 400 sec²
x = 10 ft + 0 ft + 0.5 * 2 * 400 ft
x = 10 ft + 0 ft + 0.5 * 800 ft
x = 10 ft + 0 ft + 400 ft
x = 410 ft
Therefore, the position of the car when t = 20 is 410 ft.