Car A is travelling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind Car B when the drive of car A accelerates his car with an acceleration of2.4 m/s2. How long does it take Car A to overtake Car B?
240km
58 degree south of west
19s
To solve this problem, we will use the equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
where:
s = displacement
u = initial velocity
t = time
a = acceleration
Let's assume both cars started from the same point:
For Car A:
Initial velocity (u) = 22.0 m/s
Acceleration (a) = 2.4 m/s^2
Displacement (s) = 300 m (as Car A is 300 m behind Car B)
For Car B:
Initial velocity (u) = 29.0 m/s
Acceleration (a) = 0 m/s^2 (as Car B maintains a constant velocity)
To find the time it takes for Car A to overtake Car B, we need to find the time when their displacements are equal.
Using the equation of motion for Car A:
\[ s_A = u_A t + \frac{1}{2}a_A t^2 \]
Using the equation of motion for Car B:
\[ s_B = u_B t \]
We can set these two equations equal to each other to find the time (t):
\[ u_A t + \frac{1}{2}a_A t^2 = u_B t \]
Rearranging the equation:
\[ \frac{1}{2}a_A t^2 + u_A t - u_B t = 0 \]
Now we can substitute the given values:
\[ \frac{1}{2}(2.4) t^2 + 22.0 t - 29.0 t = 0 \]
\[ 1.2 t^2 + 22.0 t - 29.0 t = 0 \]
\[ 1.2 t^2 - 7.0 t = 0 \]
Factoring out t:
\[ t(1.2 t - 7.0) = 0 \]
Setting each factor equal to zero:
\[ t = 0 \] (This is the initial time when both cars started)
\[ 1.2 t - 7.0 = 0 \]
\[ 1.2 t = 7.0 \]
\[ t = \frac{7.0}{1.2} \]
\[ t \approx 5.83 \]
Therefore, it takes approximately 5.83 seconds for Car A to overtake Car B.
To determine how long it takes Car A to overtake Car B, we need to find the time it takes for Car A to cover the distance between the two cars.
Let's assume that Car A overtakes Car B after time t, and during that time, Car B maintains a constant velocity of 29.0 m/s.
We can start by determining the distance Car A travels during time t:
Distance = Initial velocity × Time + 0.5 × Acceleration × Time^2
Since Car A is initially 300 m behind Car B, the distance Car A travels is:
Distance = 300 m + (22.0 m/s) × t + 0.5 × (2.4 m/s^2) × t^2
Now we need to set this distance equal to the distance Car B travels in time t:
29.0 m/s × t
Equating the two distances gives us the equation:
300 m + (22.0 m/s) × t + 0.5 × (2.4 m/s^2) × t^2 = 29.0 m/s × t
Simplifying this equation, we get a quadratic equation in terms of t:
0.5 × (2.4 m/s^2) × t^2 + (22.0 m/s - 29.0 m/s) × t + (300 m - 0) = 0
Now we can solve this quadratic equation to find the value of t. We can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this equation, a = 0.5 × (2.4 m/s^2), b = (22.0 m/s - 29.0 m/s), and c = (300 m - 0).
Plugging in these values, we get:
t = [-(22.0 m/s - 29.0 m/s) ± √((22.0 m/s - 29.0 m/s)^2 - 4 × 0.5 × (2.4 m/s^2) × (300 m - 0))] / 2 × 0.5 × (2.4 m/s^2)
Simplifying further will give us two values of t, one of which is the time when Car A overtakes Car B.