A bike traveling initially at a speed of 32 m/s accelerates uniformly at the rate of 3 m/s2 for a distance of 40 meters. The
bike's velocity after covering this distance is _m/s.

The  bike's velocity after covering 40 m distance is 35.55 m/s.

Explanation:

Given:

The initial velocity of the bike = 32 m/s

The rate of acceleration of the bike = 3 ms^2

Distance covered by the bike  = 40 m

To find:

The final velocity of the bike.

Solution:

The initial velocity of the bike = u = 32 m/s

The rate of acceleration of the bike = a = 3 ms^2

Distance covered by the bike = s = 40 m

The final velocity of the bike = v

Using the third equation of motion, which is written as :

The  bike's velocity after covering 40 m distance is 35.55 m/s.

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We are given:

Initial velocity (u) = 32 m/s

Acceleration (a) = 3 m/s²

Displacement (s) = 40 m

Final Velocity (v) = v m/s

Solving for the Final Velocity:

from the third equation of motion:

v² - u² = 2as

replacing the variables

v² - (32)² = 2(3)(40)

v² = 240 + 1024

v² = 1264

v = √1264

v = 35.5 m/s

Therefore, the velocity of the bike after travelling 40 m is 35.5 m/s

6 cos 34, 6 sin 34 trig

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