A race scooter is seen accelerating uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the scooter and the distance travelled.
Answers
a = v/t
= 46.1 - 18.5 / 2.47
a = 11.2 m/s^2
d = vi × t + 0.5 × a × t ^2
d = 18.5 m/s × 2.47 + 0.5 × (11.2 m/s^2) × (2.47 s)^2
d = 45.7 m + 34.1 m
d = 79.8 m
Answer:
Given ,
Initial velocity of the race scooter (u) = 18.5 m/s .
Final velocity of the race scooter (v) = 46.1 m/s .
Time (t) = 2.47 s
Acceleration = ( v - u ) / t
= A = ( 46.1 - 18.5 ) / ( 2.47 ) m/s²
= A = 27.6 / 2.47 m/s²
= A = 11.17 m/s²
The acceleration is 11.17 m/s² .
Distance travelled = ?
Distance = ut + 1/2 at²
⇒ D = 18.5 × 2.47 + 1/2 × 11.17 × (2.47)²
⇒ D = 45.695 + 34.07 m
⇒ D = 79.76 m
The distance ≈ 79.8 m.
NOTE:
By Laws of Motion :
Acceleration = ( change in velocity ) / time
⇒ Acceleration = ( final velocity - initial velocity ) / time .
The distance is denoted by S :
S = ut + 1/2 at²