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AVprozaik [17]
3 years ago
12

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).
Physics
1 answer:
DiKsa [7]3 years ago
3 0

Answer:

(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

a=\dfrac{23.5 -17.8 }{4.68}\\\\a=1.21\ m/s^2

Case 2.

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

a=\dfrac{23.5 -15.3}{4.68}\\\\a=1.75\ m/s^2

Hence, this is the required solution.

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