A race car can be slowed with a constant acceleration of -11 m/s^2
a. If the car is going 55 m/s, how many meters will it travel before it stops?
b. How many meters will it take to stop a car going twice as fast?
I'm confused a little, for these two problems they both want distance? or just time for a?
ok like one of the teachers previously stated to find the time.
V = Vi + at
0 = 55 - 11t
55 = 11t
5 = t
But for this problem i think you have to find two distance.
A, to find the distance
B, to find the distance i guess if the time is doubled, right?
A. stopping distance = (stopping time) x (average speed)
= (V/a)*(V/2) = V^2/(2a)
B. If the speed doubles, with deceleration rate "a" staying the same, the stopping distance is four times farther.
56 mph7 34mph
For both questions, we need to find the distance traveled before the car stops. The given constant acceleration of -11 m/s^2 represents the rate at which the car slows down. Therefore, we can use the equations of motion to find the distance.
a. To find the distance traveled before the car stops when it is initially going at 55 m/s, we need to find the time it takes to come to a stop. The equation we can use is:
v = u + at
where:
- v is the final velocity (0 m/s since the car stops),
- u is the initial velocity (55 m/s),
- a is the acceleration (-11 m/s^2),
- t is the time it takes to come to a stop.
Rearranging the equation, we get:
t = (v - u) / a
Substituting the values, we have:
t = (0 - 55) / (-11) = 5 seconds
Now we can use the equation for distance:
s = ut + (1/2)at^2
where:
- s is the distance,
- u is the initial velocity,
- t is the time taken,
- a is the acceleration.
Substituting the values, we have:
s = 55 * 5 + (1/2) * (-11) * (5^2)
s = 275 - 137.5
s = 137.5 meters
Therefore, the race car will travel 137.5 meters before it stops when it is initially going at 55 m/s.
b. Now, let's find the distance it takes to stop a car going twice as fast, which means the initial velocity will be 2 * 55 = 110 m/s.
Following the same process as in part a, we find the time it takes to stop:
t = (0 - 110) / (-11) = 10 seconds
Using the equation for distance:
s = 110 * 10 + (1/2) * (-11) * (10^2)
s = 1100 - 550
s = 550 meters
Therefore, it will take 550 meters to stop a car that is initially going twice as fast, at 110 m/s.