Coasting due east on your bicycle at 8.0 m/s, you encounter a sandy patch of road 7.4 m across. When you leave the sandy patch your speed has been reduced to 5.5 m/s. Assuming the bicycle slows with constant acceleration, what was its acceleration in the sandy patch? Give both magnitude and direction

a = (V^2-Vo^2)/2d

a = (5.5^2-8^2)/14.8 = -2.28 m/s^2.

To find the acceleration of the bicycle in the sandy patch, we can use the equation of motion:

v^2 = u^2 + 2as

Where:
v is the final velocity (5.5 m/s),
u is the initial velocity (8.0 m/s),
a is the acceleration in the sandy patch (what we need to find), and
s is the distance traveled in the sandy patch (7.4 m).

Rearranging the equation and solving for acceleration (a):

a = (v^2 - u^2) / (2s)

Plugging in the given values:

a = (5.5^2 - 8.0^2) / (2 * 7.4)

a = (30.25 - 64) / 14.8

a = -33.75 / 14.8

a ≈ -2.28 m/s^2

The magnitude of the acceleration in the sandy patch is approximately 2.28 m/s^2. The negative sign indicates that the acceleration is in the opposite direction of the initial motion (east).

To find the acceleration in the sandy patch, we can use the equation of motion:

v^2 = u^2 + 2as

where:
v = final velocity (5.5 m/s)
u = initial velocity (8.0 m/s)
s = distance covered (7.4 m)
a = acceleration

Rearranging the equation, we have:

a = (v^2 - u^2) / (2s)

Plugging in the given values:

a = (5.5^2 - 8.0^2) / (2 * 7.4)

a = (-4.75) / (14.8)

a ≈ -0.321 m/s^2

The magnitude of the acceleration is approximately 0.321 m/s^2, and the direction is westward (opposite to the east direction).