Answer:
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
