Mass , force and velocity are given.

Question

The driver of a 2.0 × 10³ kg car A traveling on the highway at 45m/s slams on his brakes to avoid striking a second car B in front of him, which had come to rest because of blocking. After the brakes are applied, a constant friction force of 7.5 × 10³ N acts on the car. Ignore air resistance.

(a) Determine the least distance should the brakes be applied to avoid a collision with the other vehicle?

(b) If the distance between the vehicles is initially only 40.0 m, at what speed would the collision occur?

Answer 1

(a) In order to find the distance in this case we will apply formula from kinematics

v² = 2ad,

where v = 45 m/s - speed of the car at the moment when the driver starts to apply brakes,

a - magnitude of deceleration of the car,

d - stopping distance.

From the formula the least distance

d = v² / 2a

We could find magnitude of deceleration, using the Newtons second law

a = f /m

Here

f - friction force, the only unbalanced force in this case,

m - mass of the car.

Finally we will get that

d = v²m / 2f

Plugging in all values, we will have for least distance

d = (45 m/s)²·2 x 10³ kg / 2·7.5 x 10³ N = 270 m

(b) if the initial distance between the cars d = 40.0 m, then the collision speed will be

v = √ 2fd / m

After substitution of all values, we obtain

v = √ 2·7.5 x 10³ N·40.0 m / 2 x 10³kg = 17.3 m/s