What will the stopping distance be for a 1000kg car if 2000 N of force are applied when the car is traveling 10 ms?

Answer 1

This is clearly a case of motion under constant acceleration(deceleration).

The equation you need to use would be:

(1) S = vo - 0.5 * a * t2, where:

S - distance,

vo - starting velocity,

a - acceleration,

t - time.

Acceleration can be easily found from

(2) a = F / m, where:

F - force(here: 2000 N),

m - mass(of the car: 1000 kg).

Time it takes for a car to stop is a time it takes for its starting velocity to drop to 0 with a constant rate of deceleration(a):

(3) t = vo / a

Substituting (2) and (3) into (1), we get:

S = vo - 0.5 * (F / m) * vo2 / (F / m)2, which simplifies to:

(4) S = vo - 0.5 * mvo2 / F.

Now, after substituting numerical values, we get our answer:

S = 10[m/s] - 0.5 * 1000[kg] * 100[m2/s2] / 2000 N =

= 25[m]

Note: 1[N] = 1[kg * m / s2], which makes above equation with units check out.

Answer 2

To calculate stopping distance, you need to determine the deceleration first. Using Newton's second law (Force = mass x acceleration), you can find that the deceleration is 2 m/s^2. Then, you can use the equation of motion (stopping distance = initial velocity^2 / (2 x deceleration)) to find the stopping distance, which in this case would be 25 meters.