Car A travelling at 13 m/s [N] is 190m behind car B which is travelling at 9.0 m/s [N]. At this moment car A starts to accelerate at 6.0 m/s^2 [N] to catch up with car B. After 1.0s, car B also starts to accelerate at 2 m/s^2 [N].
How long will it take car A to catch up with car B?
To calculate the time it will take for Car A to catch up with Car B, we need to determine the positions of both cars at any given time.
Let's start by finding the position of Car A after 1.0 second. Given that Car A is initially 190m behind Car B and both cars are moving north (as indicated by the [N]), we can use the equation:
Position (d) = Initial position (d0) + (Initial velocity (v0) × time (t)) + (0.5 × acceleration (a) × time^2)
For Car A:
Initial position (d0) = -190m (negative because it is behind Car B)
Initial velocity (v0) = 13 m/s (given)
Acceleration (a) = 6 m/s^2 (given)
Time (t) = 1.0s
Plugging in the values, we have:
d_A = -190m + (13 m/s × 1.0s) + (0.5 × 6 m/s^2 × (1.0s)^2)
Simplifying the equation:
d_A = -190m + 13 m + 0.5 × 6 m/s^2 × 1.0s^2
d_A = -190m + 13m + 0.5 × 6 m/s^2
d_A = -190m + 13m + 3 m
d_A = -174m
So, after 1.0 second, Car A is at a distance of -174m (which means it is still behind Car B).
Next, let's calculate the position of Car B after 1.0 second, considering its initial position, velocity, and acceleration:
For Car B:
Initial position (d0) = 0m (assumed, as it is not mentioned)
Initial velocity (v0) = 9.0 m/s (given)
Acceleration (a) = 2.0 m/s^2 (given)
Time (t) = 1.0s (given)
Using the same equation as above:
d_B = 0m + (9.0 m/s × 1.0s) + (0.5 × 2.0 m/s^2 × (1.0s)^2)
Simplifying the equation:
d_B = 0m + 9.0 m + 0.5 × 2.0 m/s^2
d_B = 0m + 9.0m + 1.0 m
d_B = 10.0m
Therefore, after 1.0 second, Car B has moved ahead to a position of 10.0m.
To determine when Car A catches up with Car B, we need to find the time at which their positions are equal. This can be done by solving the equation:
d_A + 190m = d_B
-174m + 190m = 10.0m
16m = 10.0m
Combining the distances:
16m - 10.0m = 6.0m
We know that Cars A and B both accelerate from that point forward, so we can set up the following equation to find the time it takes for Car A to catch up:
d_A + (velocity of A × time) + (0.5 × acceleration of A × time^2) = d_B + (velocity of B × time) + (0.5 × acceleration of B × time^2)
Plugging in the known values:
-174m + (13 m/s × t) + (0.5 × 6 m/s^2 × t^2) = 10.0m + (9.0 m/s × t) + (0.5 × 2 m/s^2 × t^2)
Combining like terms:
-174m + 13 m/s × t + 3 m/s^2 × t^2 = 10.0m + 9.0 m/s × t + 1.0 m/s^2 × t^2
Rearranging the equation:
3 m/s^2 × t^2 - 1.0 m/s^2 × t^2 - 9.0 m/s × t + 13 m/s × t - 10.0m + 174m = 0
2.0 m/s^2 × t^2 + 4.0 m/s × t + 164m = 0
Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
a = 2.0 m/s^2
b = 4.0 m/s
c = 164m
t = (-4.0 m/s ± sqrt((4.0 m/s)^2 - 4 × 2.0 m/s^2 × 164m)) / (2 × 2.0 m/s^2)
Simplifying the equation:
t = (-4.0 m/s ± sqrt(16 m^2/s^2 - 1312 m^2/s^2)) / 4.0 m/s^2
t = (-4.0 m/s ± sqrt(-1296 m^2/s^2)) / 4.0 m/s^2
Since the square root of a negative number does not produce a real solution, it means that Car A will not catch up with Car B in this scenario.