Review Conceptual Example 6 as background for this problem. A car is traveling to the left, which is the negative direction. The direction of travel remains the same throughout this problem. The car's initial speed is 17.8 m/s, and during a 4.68-second interval, it changes to a final speed of (a)23.5 m/s and (b)15.3 m/s. In each case, find the acceleration (magnitude and algebraic sign).

Acceleration is the change in the velocity of a body with time.

The term acceleration is the change in the velocity of a body with time. We have to find the acceleration in two separate cases;

Case 1;

a = -(23.5 m/s - 17.8 m/s)/4.68-s

a = -1.2 m/s^2

Case 2;

a = -(15.3 m/s - 17.8 m/s)/4.68-s

a = 0.53 ms-2

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(a) 1.21 m/s² (b) 1.75 m/s²

Explanation:

The initial speed of the car, u = 17.8 m/s

Case 1.

Final speed of the car, v = 23.5 m/s

Time, t = 4.68-s

Acceleration = rate of change of velocity

Case 2.

Final speed of the car, v = 15.3 m/s

Hence, this is the required solution.

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2