Question

If a cat walks at 1.3 m/s and accelerates at 3.2 m/s^2, how far does it travel after 8.2 seconds?

a. 29.1 m
b. 33.2 m
c. 42.1 m
d. 54.4 m

Answer 1

When using the kinematic equation s = ut + ½at² to calculate the distance a cat travels after accelerating for 8.2 seconds at a rate of 3.2 m/s² from an initial speed of 1.3 m/s, the result is approximately 118.148 meters.

Step-by-step explanation:

To calculate the distance a cat travels after it accelerates for 8.2 seconds at an acceleration rate of 3.2 m/s² from an initial speed of 1.3 m/s, we use the kinematic equation: s = ut + ½at², where:s is the displacement (distance traveled), u is the initial velocity (1.3 m/s), t is the time (8.2 seconds), a is the acceleration (3.2 m/s²). Plugging the values into the equation, we get: s = (1.3 m/s)(8.2 s) + ½(3.2 m/s²)(8.2 s)², s = 10.66 m + 107.488 m, s = 118.148 m. The cat travels a distance of approximately 118.148 meters, which is not one of the options provided in the question. Hence, if the options provided in the question are to be considered, there may have been a mistake either in the options or in the given values for the calculation.