The graph shows the motion of a cyclist. Calculate;

a) The acceleration in the first 20 s.

b) The acceleration between 20 and 30s.

c) The acceleration in the last 10 s.

d) The distance travelled by the cyclist when he was moving at a constant speed?
Show working.

A. 0.5m/s².

B. –0.5m/s².

C. –0.5m/s².

D. 100m.

Explanation:

A. Determination of the acceleration in the first 20s.

Initial velocity (u) = 0

Final Velocity (v) = 10m/s

Time (t) = 20secs.

Acceleration (a) =..?

Acceleration = change in Velocity /time

a = (v – u)/t

a = (10 – 0)/20 = 10/20

a = 0.5m/s²

Therefore, the acceleration of the cyclist in the first 20secs is 0.5m/s²

B. Determination of the acceleration between 20 and 30s. This can be obtained as follow:

Initial velocity (u) = 10m/s

Final Velocity (v) = 5m/s

Time (t) = 30 – 20 = 10s

Acceleration (a) =..?

Acceleration = change in Velocity /time

a = (v – u)/t

a = (5 – 10)/10 = –5/10

a = –0.5m/s²

Therefore, the acceleration of the cyclist between the 20 and 30secs is

–0.5m/s².

C. Determination of the acceleration in the last 10s.

Initial velocity (u) = 5m/s

Final Velocity (v) = 0

Time (t) = 10s

Acceleration (a) =..?

Acceleration = change in Velocity /time

a = (v – u)/t

a = (0 – 5)/10 = –5/10

a = –0.5m/s²

Therefore, the acceleration of the cyclist between the last 10secs is

–0.5m/s².

D. Determination of the distance travelled by the cyclist when he was moving at a constant speed.

Velocity (v) = 5m/s

Time (t) = 50 – 30 = 20secs

Displacement (d) =?

Velocity = Displacement /Time

v = d/t

5 = d/20

Cross multiply

d = 5 x 20

d = 100m

Therefore, the distance travelled by the cyclist at constant speed is 100m

the answer is c. the lower legs are levers, and the knees are fulcrums. the ankles hold the loads.