Question

1. The velocity of an object traveling along the x-axis changes from 20 m/s to 60 m/s in 10 seconds. What was the displacement that occurred during this time? 2. An object moving along the x-axis initially has a velocity of 60 m/s and is slowing rest? down at the rate of 4 m/s per second. How far (in meters) will it travel before coming to 3. An accelerating object travels 12 meters from rest in a certain time. How far (in meters) could it have traveled from rest in triple the time? 4. An object's initial velocity is 20 m/s and is accelerating. When the object has traveled 240 m, its velocity is 80 m/s. What is the object's acceleration (in m/s)? 5. Two cars initially separated by 1000 m are traveling away from each other. Car A is moving at 20 m/s and is accelerating at 4 m/s. Car B is moving at 10 m/s and is accelerating at 6 m/s. About how far (in meters) will Car A have traveled when the cars are 4000 meters apart? 6. A sailboat leaves a harbor and sails 1.8 km in the direction 65° south of east, where the captain stops for lunch. A short time later, the boat sails 1.1 km in the direction 15° north of east. What is the magnitude of the resultant displacement from the harbor?

Answer 1

1.

Using 1st kinematic equation:

V = U + at

a = (V - U)/t

V = final speed = 60 m/sec

U = initial speed = 20 m/sec

t = time = 10 sec

So,

a = (60 - 20)/10 = 4 m/sec^2

Now Using this value in 2nd kinematic equation

S = U*t + 0.5*a*t^2

S = 20*10 + 0.5*4*10^2

S = 400 m = displacement during this time

2.

Using 3rd kinematic equation

V^2 = U^2 + 2*a*S

S = Stopping distance = (V^2 - U^2)/(2*a)

V = final speed = 0 m/sec

U = Initial speed = 60 m/sec

a = de-acceleration = -4 m/sec^2

So,

S = (0^2 - 60^2)/(2*(-4))

S = 450 m

3.

Using 2nd kinematic equation

S = U*t + 0.5*a*t^2

given that it travels 12 m in time 't'

U = initial speed = 0 m/sec

12 = 0*t + 0.5*a*t^2

a = 12/(0.5*t^2) = 24/t^2

Now when time is '3t' distance traveled will be

S1 = U*t1 + 0.5*a*t1^2

S1 = 0*3t + 0.5*(24/t^2)*(3t)^2

S1= 0 + 12*9

S1 = 108 m = distance traveled during triple time

4.

from 3rd kinematic equation:

a = (V^2 - U^2)/(2*S)

Using given values:

a = (80^2 - 20^2)/(2*240)

a = 12.5 m/sec^2 = acceleration of object

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