High School

A driver of a car traveling at -15 m/s applies the brakes, causing a uniform acceleration of +2.0 m/s\(^2\).

1. If the brakes are applied for 2.5 s, what is the velocity of the car at the end of the braking period?
2. How far has the car moved during the braking period?

Answer :

Given :

Initial velocity, u = -15 m/s.

Acceleration , a = 2 m/s².

Time taken to applied brake, t = 2.5 s.

To Find :

The velocity of the car at the end of the braking period.

How far has the car moved during the braking period.

Solution :

By equation :

[tex]v = u+at\\\\v=-15 + 2\times 2.5\\\\v=-10 \ m/s[/tex]

Now, distance covered by car is :

[tex]s=ut+\dfrac{at^2}{2}\\\\s=(-15)(2.5)+\dfrac{2(2.5)^2}{2}\\\\s=-31.25\ m[/tex]

Hence, this is the required solution.

The final velocity of the car after braking for 2.5 seconds with an initial velocity of -15 m/s and a uniform acceleration of +2.0 m/s² is -10 m/s. During this time, the car covers a distance of -31.25 meters.

The question involves calculating the final velocity of a car after braking and the distance covered during the braking period. To find the final velocity, we use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

The car's initial velocity is -15 m/s (the negative sign indicates direction), the acceleration is +2.0 m/s², and the brakes are applied for 2.5 s. Plugging the values into the equation we get: v = (-15 m/s) + (+2.0 m/s²)(2.5 s) = -10 m/s.

To find the distance traveled during the braking period, we can use the equation s = ut + (1/2)at², where s is the distance.

Substituting the known values, we get: s = (-15 m/s)(2.5 s) + (1/2)(+2.0 m/s²)(2.5 s)² = -37.5 m + 6.25 m = -31.25 m.

The negative sign indicates that the displacement is in the opposite direction to the initial velocity.