Final answer:
The car's velocity after 2.50 seconds will be 12.5 m/s, it will have moved 42.1875 m during this time, and it will take 8.0 seconds in total for the car to come to a complete stop.
Step-by-step explanation:
Understanding Car Braking Dynamics Using Physics
To answer these questions about a car's velocity and distance travelled while braking, we'll use the principles of physics, specifically the impulse-momentum theorem and kinematic equations.
a. Car's velocity after 2.50s:
The impulse-momentum theorem states that the change in momentum (∆p) of an object is equal to the impulse (J) applied to it. Impulse is the product of the force (F) and the time (t) for which it acts, J = F × t.
Given:
Force (F) = -6250 N (negative since it's opposite to the direction of motion),
Time (t) = 2.50 s,
Initial velocity (u) = 20.0 m/s,
Mass (m) = 2500 kg.
The acceleration can be found using Newton's second law, F = m × a. So, a = F / m = -6250 N / 2500 kg = -2.5 m/s².
The final velocity (v) after 2.50s can be found using the kinematic equation v = u + a × t, thus v = 20.0 m/s + (-2.5 m/s²) × (2.50 s) = 12.5 m/s.
b. Distance moved during 2.50 s:
We can use the kinematic equation s = ut + ½at² to calculate the distance (s). Plugging in the values, we get s = (20.0 m/s) × (2.50 s) + ½ × (-2.5 m/s²) × (2.50 s)² = 50.0 m - 7.8125 m = 42.1875 m.
c. Time taken to come to a complete stop:
Using the formula v = u + at again and setting v to 0 for a complete stop, we can solve for t: 0 = 20.0 m/s + (-2.5 m/s²) × t. Rearranging, we get t = -20.0 m/s / -2.5 m/s² = 8.0 s.
The car will take 8.0 seconds to come to a complete stop under these conditions.