Calculate the displacement and velocity at times of (a)...

Name: Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.
Uploaded: 2022-05-07T19:48:04-08:00
Duration: 6 min 53 s
Channel: Sikandar Baig
Description: Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Transcript

00:01 Dear student, so in path a, let's start from part a, initial velocity is given.

00:15 Excellation due to height is well known, height of the bridge is given.

00:19 So for the displacement, we can use equation of motion.

00:22 So equation of motion is the total distance is equal to the product of initial velocity times t plus half times acceleration into the.

00:33 Square of time all right so let's plug in the values the initial velocity is minus 14 meters per second time is 0 .5 seconds i'm skipping the units because all the units are in si in si system half acceleration is negative 9 .8 meters per second square time is 0 .5 seconds and it's scared so from here the total distance comes out to be 9 .45 meters it is negative it is against the gravitational acceleration now the final velocity should be equal to the sum of initial velocity plus the product of acceleration time let's plug in the values the initial velocity is minus 14 meters per second plus the acceleration is 9 .8 meters per second time is half second so from here the final velocity comes out to be negative 18 .9 meters per second i hope you understood this so let's move to part b here again we will use the same formula the total distance is equal to the product of initial velocity with time half times acceleration into the scale of time so let's plug in the values initial velocity is minus 40 here the time is one second plus half acceleration is same negative 9 .8 meters per second time is one second and its whole square.

02:29 So from here the total distance comes out to be minus 18 .9 meters.

02:38 Same we will do.

02:40 Final velocity is equal to the initial velocity plus the product of acceleration and time.

02:48 So the initial velocity is negative 14 meters per second plus the acceleration is negative 9 .8 meters per second square.

02:57 Time is one second year.

03:00 So date equals minus 23 .8 meters per second.

03:07 All right, dear students.

03:09 Now let's move to part c...