Two motorcars, A and B are approaching each other on a highway (Figure 3). At time t = 0, car A and B are 975 m apart at point P and Q, with their speeds being uA = 30 m/s and uB = 17 m/s, respectively. Car A passes point Q after 40 seconds and car B passes point P after 42 seconds.

Question

Find their accelerations

Answer 1

To find the accelerations of the two cars, we first need to find their positions at the time they pass each other.

Let x be the position of car A at time t, and y be the position of car B at time t. Then we have:

x = 975 m + 30 m/s * t
y = 975 m - 17 m/s * t

We want to find the time t when car A passes point Q and when car B passes point P.

For car A passing point Q:
975 m + 30 m/s * t = 0
30 m/s * t = -975 m
t = -32.5 s (ignore the negative sign as time cannot be negative)

For car B passing point P:
975 m - 17 m/s * t = 0
17 m/s * t = 975 m
t = 57.35 s

Now, we can find the positions of the cars at these times:

x = 975 m + 30 m/s * 32.5 s
x = 975 m + 975 m
x = 1950 m

y = 975 m - 17 m/s * 57.35 s
y = 975 m - 975 m
y = 0 m

So, car A passes car B at a distance of 1950 m from point P.

Now, we can find the accelerations of the two cars:

Acceleration of car A:
vA = uA + aA * t
0 = 30 m/s + aA * 32.5 s
aA = -0.92 m/s^2

Acceleration of car B:
vB = uB + aB * t
0 = 17 m/s + aB * 57.35 s
aB = -0.296 m/s^2

Therefore, the accelerations of car A and car B are -0.92 m/s^2 and -0.296 m/s^2, respectively.