a car traveling at 28 m/s slows down at a constant rate for 4 seconds until it stops. what is its acceleration?
1 answer:
We are given the following conditions:
Since we are told it slows down at a constant rate, we know we are dealing with uniformed accelerated motion, or UAM for short.
So we look at our kinematics equation that contains the given variables.
This contains our target variable and our initial conditions. Plug and chug, we get:
Thus the answer is:
a = -7m/s^2
Hope this helps!
Since we are told it slows down at a constant rate, we know we are dealing with uniformed accelerated motion, or UAM for short.
So we look at our kinematics equation that contains the given variables.
This contains our target variable and our initial conditions. Plug and chug, we get:
Thus the answer is:
a = -7m/s^2
Hope this helps!
You might be interested in
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
where k is the elastic constant of the spring and
is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
where the negative sign is given by the fact that
points in the opposite direction of the displacement of the cart, and where
therefore, the work done by the weight is
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
where k is the elastic constant of the spring and
(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
where the negative sign is given by the fact that
therefore, the work done by the weight is