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.
Answer:
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
:


I believe it’s A. I know for sure it isn’t D.
Answer:
this is impossible for me
Explanation:
Answer:
Explanation:
Given
time taken 
Speed acquired in 2 sec 
Here initial velocity is zero 
acceleration is the rate of change of velocity in a given time


Distance travel in this time

where
s=displacement
u=initial velocity
a=acceleration
t=time


so Jet Plane travels a distance of 42 m in 2 s