Question

What will the stopping distance be for a 2,000-kg car if -2,000 N of force are applied when the car is traveling 20 m/s?

Answer 1

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 :      

$F=ma$

$a=\dfrac{F}{m}$

$a=\dfrac{-2000}{2000}$

$a=-1\ m/s^2$

Let s is the stopping distance. Now using third equation of motion as :

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{0-(20)^2}{2\times -1}$

s = 200 meters

So, the stopping distance of the car is 200 meters. Hence, this is the required solution.

Answer 2

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