a race car can be slowed with a constant acceleration of -11m/s2.
a.if the car is going 55m/s,how many meters will it travel beofre it stops?
b.how many metes will it take to stop a car going twice as fast?
it takes 5 s to stop the car __ 55 / 11
the average speed (Va) is __ (55 + 0) / 2
the stopping distance is __ Va * 5
twice as fast means that the stopping time is doubled, along with the average speed
__ this means that the car travels 4 times as far
To answer these questions, we can use the kinematic equation that relates the change in velocity (Δv), acceleration (a), and distance (d):
v^2 = u^2 + 2ad
Where:
- v is the final velocity.
- u is the initial velocity.
- a is the acceleration.
- d is the distance.
Let's solve each question step by step:
a. If the car is going 55 m/s, we need to find how many meters it will travel before it stops.
Given data:
- Initial velocity (u) = 55 m/s.
- Acceleration (a) = -11 m/s^2 (negative value indicates deceleration).
- Final velocity (v) = 0 m/s (since the car stops).
By substituting the values into the equation, we get:
0^2 = 55^2 + 2(-11)d
Simplifying the equation:
0 = 3025 - 22d
Rearranging the equation to isolate 'd':
22d = 3025
d = 3025 / 22
Therefore, the car will travel approximately 137.5 meters before it stops.
b. Now, let's find out how many meters it will take to stop a car going twice as fast. If the initial velocity (u) is twice as fast as before, then u = 2 * 55 = 110 m/s.
Using the same equation:
0^2 = 110^2 + 2(-11)d
Simplifying:
0 = 12100 - 22d
Rearranging:
22d = 12100
d = 12100 / 22
The car will travel approximately 550 meters before it stops.