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.
