A body moving in the positive x direction passes the origin at time t = 0. Between t = 0 and t = 1 second, the body has a constant speed of 24 meters per second. At t = 1 second, the body is given a constant acceleration of 6 meters per second squared in the negative x direction. The position x of the body at t = 11 seconds is

-99? I think I did this wrong

Answer: Because your acceleration from t=1 to t=11 is 6 meters per second squared. All you need to do is integrate 6 to get the velocity. You would get 6x+c. Ignoring the constant, just multiply 6 by 10 seconds. 6(10) = 60, you get that for a body with a constant acceleration of 6 meters per second squared, in 10 seconds, it will travel a total of 60 meters. However, since this is in the negative-x direction, you must subtract this from the 24 meters in which the body already traveled in the 1st second. 24-60, you will get -36m.

so I tried plugging 11 into that x= 24 + 24t - (1/2) t^2 and I got -75. Would that be the right answer?

I suspect that the problem writer used 11 instead of 10 seconds for the second part, but that is wrong. Choose -75

Well, you're not entirely wrong. Let's see if we can figure it out together.

Since the body has a constant speed of 24 meters per second between t = 0 and t = 1 second, we know that its position at t = 1 second is 24 meters.

Now, to find the position at t = 11 seconds, we need to determine the distance the body moves during the time interval from t = 1 to t = 11 seconds.

To do this, we first need to find the time it takes for the body to stop moving. Since the body has a constant negative acceleration of 6 meters per second squared, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

In this case, the initial velocity (u) is 24 meters per second, the final velocity (v) is 0 meters per second (since the body comes to a stop), and the acceleration (a) is -6 meters per second squared.

Using the equation v = u + at, we can rearrange it to solve for t:

t = (v - u) / a
t = (0 - 24) / (-6)
t = 24 / 6
t = 4 seconds

So, it takes 4 seconds for the body to stop moving. Now, let's find out how far it travels during this time.

We can use the equation s = ut + (1/2)at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time taken.

In this case, the initial velocity (u) is 24 meters per second, the acceleration (a) is -6 meters per second squared, and the time taken (t) is 4 seconds.

Plugging in the values, we have:

s = (24 * 4) + (1/2) * (-6) * (4^2)
s = 96 + (-6) * 16
s = 96 + (-96)
s = 0

So, during the 4 seconds, the body doesn't actually travel any distance. It comes to a stop at the original position (as we calculated earlier), and then stays there for the remaining time.

Therefore, the position x of the body at t = 11 seconds is the same as the position at t = 1 second, which is 24 meters.

Don't worry, you did great! Keep up the good work!

To find the position of the body at t = 11 seconds, we can calculate it using the equations of motion. We will need to break down the problem into two parts: the initial motion at constant speed and the subsequent motion with constant acceleration.

First, let's calculate the position of the body during the initial motion when it has a constant speed of 24 m/s. We can use the formula:

x = x0 + v0 * t

Where:
x = final position of the body,
x0 = initial position of the body,
v0 = initial velocity of the body, and
t = time

Given that the body passes the origin at t = 0 with a positive x direction, we can set x0 = 0 and v0 = 24 m/s. Plugging in the values, we get:

x1 = 0 + 24 * 1 = 24 meters

So, at t = 1 second, the body's position is 24 meters.

Next, let's calculate the position of the body during the time when it has constant acceleration. We can use the formula:

x = x0 + v0 * t + (1/2) * a * t^2

Where:
a = acceleration

Given that the body starts with an initial velocity of 24 m/s and experiences a constant acceleration of -6 m/s^2, we can set x0 = 24 meters, v0 = 24 m/s, a = -6 m/s^2, and t = 11 - 1 = 10 seconds. Plugging in the values, we get:

x2 = 24 + 24 * 10 + (1/2) * (-6) * (10^2)
= 24 + 240 - 300
= -36 meters

So, at t = 11 seconds, the body's position is -36 meters.

Therefore, the position x of the body at t = 11 seconds is -36 meters.

at t = 1 it is at x = 24

a = -6
so now start over for 10 seconds to get to t = 11
at new time = 0, Vi = +24 and a = -6
v = 24 - 6 t
x = 24 + 24 t - (1/2)6 t^2
at t = 10
x = 24*10 - 3 (100)
240 - 300 is indeed not -99

ok now I'm even more confused. So the answer is -60?

But the options for the question are
a)- 99 m
b)+ 99 m
c)+ 36 m
d)- 75 m
e)- 36 m