Answer :
The cart's increased speed is the same as when it first took off.
Rate is the speed from which an object travels along a path over time, whereas velocity is the speed and orientation of an item's motion.
At first, the cart is at rest.
So, u = 0 m/s.
A brief period of time is spent with a constant force applied.
The final speed that the force gives the cart is what we'll call v.
Newton's first equation of motion yields the following results:
v = u + at
Where;
The ultimate speed =v.
The starting speed =u.
Acceleration = a.
Time = t
u = 0 m/s, therefore we now have;
v = 0 + at
v = at
Let's now use Newton's second rule of motion to construct a formula to introduce force;
F = ma
Where;
Force = f.
Mass = m.
Acceleration = a.
So,
= a = F/m.
Substituting in v = at,
= v = (F/m)t
The final speed, v, is evidently exactly related to the force. So, if the force is constant, the end speed will also stay constant.
To learn more about Speed:
brainly.com/question/13232071
#SPJ4
The complete question is -
A constant force is exerted for a short time interval on a cart that is initially at rest on an air track. This force gives the cart a certain final speed. Suppose we repeat the experiment but, instead of starting from rest, the cart is already moving with constant speed in the direction of the force at the moment we begin to apply the force.
After we exert the same constant force for the same short time interval, the increase in the cart's speed:
A. is equal to two times its initial speed.
B. is equal to the square of its initial speed.
C. is equal to four times its initial speed.
D. is the same as when it started from rest.
E. cannot be determined from the information provided.
