A student walking home at a steady rate of 4.80 m/s suddenly decides to start running. She can accelerate at 7.50 m/s22, and runs for 1.70 s before stopping. What distance did the student cover while running?

in this question we see that the only movement ( displacement/distance) that could be done was horizontal distance we started off with 4.80 m/s which was the initial velocity that the person was walking by, the person was walking by this velocity in a constant rate e then the person starts to accelerate by 7.50m/s^2 and the person was able to speed up by this acceleration by running for 1.80s then the person stops we can conclude from this that the final velocity is equal to zero then we can use the formula v^2 = v1^2 +2ad and solve for d hope it helps sorry not sure if this is the right answer

To find the distance the student covered while running, we can break the problem into two parts: the time she was walking and the time she was running.

First, let's calculate the distance she covered while walking. We know her walking speed is 4.80 m/s and the total time she spent running is 1.70 s. The distance covered while walking is given by:

Distance_walked = Walking_speed * Time_walked

Distance_walked = 4.80 m/s * 1.70 s

Next, let's calculate the distance she covered while running. We know her initial speed is 4.80 m/s, acceleration is 7.50 m/s^2, and the time she spent running is 1.70 s. The distance covered while running is given by the kinematic equation:

Distance_run = Initial_speed * Time_run + (1/2) * Acceleration * (Time_run)^2

Distance_run = 4.80 m/s * 1.70 s + (1/2) * 7.50 m/s^2 * (1.70 s)^2

Now, we can calculate the total distance she covered by adding the distances covered while walking and running:

Total_distance = Distance_walked + Distance_run

Substituting the calculated values,

Total_distance = 4.80 m/s * 1.70 s + (1/2) * 7.50 m/s^2 * (1.70 s)^2

By evaluating this expression, we get the distance covered by the student while running.