Question

2. The skid marks made by an automobile indicated that its breaks were first applied for a distance of 60 meters (m) before it came to a complete stop. The car in question is known to have a constant deceleration of 20 m/s2 under these conditions. How fast (in km/h) was the car traveling when the brakes were first applied? [Be sure to sketch your coordinate system to clarify the meaning of the functions involved.] Be sure to include units in your answer.

Answer 1

The initial velocity of the car when the brakes were first applied is approximately 48.99 m/s or 176.36 km/h.

Step-by-step explanation:

To find the initial velocity of the car when the brakes were first applied, we can use the equation: v² = u² + 2as, where v is the final velocity (0 m/s in this case), u is the initial velocity (what we want to find), a is the deceleration, and s is the distance traveled. Rearranging the equation gives us: u² = v² - 2as. Plugging in the given values, we have: u² = 0² - 2(20)(60), which simplifies to: u² = -2400. Taking the square root of both sides gives us: u = - 48.99 m/s. However, since we're dealing with speed, we take the positive value, so the car was traveling at approximately 48.99 m/s when the brakes were applied.

To convert this speed to km/h, we use the conversion factor 3.6, since 1 meter per second is equal to 3.6 kilometers per hour. Thus, the car was traveling at approximately 176.36 km/h when the brakes were first applied.

Answer 2

Step-by-step explanation:

We shall take motion along x - axis ( positive )

We shall apply the equation of law of motion as follows

v² = u² - 2aS

v is final velocity , u is initial velocity , S is displacement and a is deceleration.

0 = u² - 2 x 20 x 60

u² = 2 x 20 x 60

u² = 2400

u = 49 m /s approx.

= 49 x 60 x 60 / 1000 km / h

= 176.4 km / h .