Answer is B- 200 m
Given:
m (mass of the car) = 2000 Kg
F = -2000 N
u(initial velocity)= 20 m/s.
v(final velocity)= 0.
Now we know that
F= ma
Where F is the force exerted on the object
m is the mass of the object
a is the acceleration of the object
Substituting the given values
-2000 = 2000 × a
a = -1 m/s∧2
Consider the equation
v=u +at
where v is the initial velocity
u is the initial velocity
a is the acceleration
t is the time
0= 20 -t
t=20 secs
s = ut +1/2(at∧2)
where s is the displacement of the object
u is the initial velocity
t is the time
v is the final velocity
a is the acceleration
s= 20 ×20 +(-1×20×20)/2
s= 200 m
Answer:
Stopping distance, s = 200 meters
Explanation:
Mass of the car, m = 2000 kg
Force acting in the car, F = -2000 N
Initial speed of car, u = 20 m/s
Finally, it stops, v = 0
Using second equation of motion as :
[tex]F=ma[/tex]
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{-2000}{2000}[/tex]
[tex]a=-1\ m/s^2[/tex]
Let s is the stopping distance. Now using third equation of motion as :
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{0-(20)^2}{2\times -1}[/tex]
s = 200 meters
So, the stopping distance of the car is 200 meters. Hence, this is the required solution.
