Question

At time t=0 , a cart is at x=10 m and has a velocity of 3 m/s in the −x -direction. The cart has a constant acceleration in the +x -direction with magnitude 3 m/s^2 < a < 6  m/s^2 . Which of the following gives the possible range of the position of the cart at t=1 s ?

Answer 1

Answer:

Explanation:

The minimum magnitude of acceleration = 3 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 3 x 1²

= - 3 + 1.5

= - 1.5 m

position at t = 1 s

= 10 - 1.5

= 8.5 m

The maximum  magnitude of acceleration = 6 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 6 x 1²

= - 3 + 3

=  0

position at t = 1 s

= 10 +0

= 10  m

So range of position is 8.5 m to 10 m .

Answer 2

We want to find the range of possible positions for the cart at t = 1s.

The range is:

14.5m ≤ p ≤ 16m

Finding the cart's motion equations.

First, we know that the acceleration of the cart is a, a constant (but we don't know the exact value yet) so we write the acceleration as:

a(t) = a

To get the velocity equation we integrate over time, and we know that the velocity at t = 0s is 3m/s, so that will be the constant of integration.

v(t) = a*t + 3m/s

To get the position we integrate again, the initial position is x = 10m, so that will be the constant of integration.

p(t) = (a/2)*t^2 + (3m/s)*t + 10m

Now, to get the range of possible positions for t = 1s we need to use the minimum and maximum accelerations and evaluate the position equation in t = 1s.

The minimum acceleration is a = 3m/s^2, so we have the minimum position:

p(1s) = (3m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 14.5m

The maximum acceleration is 6m/s^2, so the maximum position is:

P(1s) =  (6m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 16m

So the range of the position at t = 1s is:

14.5m ≤ p ≤ 16m

If you want to learn more about motion equations, you can read:

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