A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s. Then it slows down at a steady rate of 2.0 m/s2 for 4.0 s. What is the final speed of the car?

A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. The final speed of the car is 12 m/s.

The velocity of an object is usually referred to as the change rate of the object in a specified direction.

Given that:

the initial acceleration of the car is = 4.0 m/s²time = 5.0s

Using the first equation of motion;

Let v₁ be the initial speed of the car

v₁ = u₁ + a₁t₁

Since the car starts from rest, the initial speed = 0 m/s

v = 0 + (4.0 m/s²) × 5.0 s

v = 0 + 20 m/s

v = 20 m/s

The final acceleration = -2.0 m/s²  (since the acceleration is decelerating)time = 4.0 s

Using the same first equation of motion;

Let v₂ be the final speed of the car

v₂ = u₂ + a₂t₂

v₂ = 20 m/s + (-2 m/s²) × (4.0 s)

v₂ = (20 - 8) m/s

v₂ = 12 m/s

Therefore, we can conclude that the final speed of the car is 12 m/s.

Learn more about the speed of an object here:

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The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: where m is the mass of the arrow and v its velocity.  all the other options, instead, do not increase the kinetic energy of the system: -  pulling farther back on the string : this will just increase the elastic potential energy of the system-  holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)-  decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)