Answer:
It takes the car 2.5 s to slow down to a final speed of 10.0 m/s.
The car has moved 31.25 m during the braking period.
Explanation:
Assuming that motion is a straight line and knowing that the accelaration is a constant (-2.0 m/s²), these equations apply:
v(t) = v0 +a*t
X(t) = X0 + v0*t + 1/2*a*t²
where:
v(t) = velocity at time "t".
v0 = initial velocity.
a = acceleration.
t = time.
X(t) = position at time "t"
X0 = initial position
Given data:
v0 = 15.0 m/s
a = 2.0 m/s²
v(t) = 10.0 m/s
We need to find "t" at which the speed is 10 m/s.
From the equation of velocity above:
v(t) = v0 + a*t
solving the equation for "t"
v(t) - v0 = a*t
(v(t) - v0)/a = t
(10 m/s - 15 m/s) / (-2m/s²) = t
2.5 s = t
Now that we know how long it takes to slow down to a speed of 10 m/s, we can calculate how far has the car moved in that period using the equation for position at a given time:
X(t) = X0 + v0*t + 1/2*a*t²
Since we only want to know how many meters has the car moved during 2.5 s with an acceleration of -2m/s², we can make X0(initial position) = 0.
Then:
X(2.5 s) = 0 m + 15 m/s * 2.5 s + 1/2 * (-2m/s²)* (2.5 s)²
X(2.5 s) = 37.5 m + (-6.25 m) = 31.25 m
