The action of two forces. one is a forward force of 1157 n provided by traction between the wheels and the road. the other is a 902 n resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 m/s?

Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

(1)

with m the mass , a the acceleration and the net force forces that is:

(2)

with F the force provided by traction and f the resistive force:

(2) on (1):

solving for a:

Now let's use the Galileo’s kinematic equation

(3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and the time took to achieve that velocity, solving (3) for :

emily ruth's annual premium is $1058.50

step-by-step explanation:

emily ruth is a female of 27 years age purchasing 20 year endowment insurance with face value $25000.

from the given table,

for females, the age of a male 3 years younger should be used.

so, we have to check for 24 year male for 20 year endowment.

it is 42.34 for every $1000 face value.

for a face value of $25000,

it is 25 * 42.34 = $1058.50

∴ emily ruth's annual premium is $1058.50

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft