From my experience, I would say it is true.
The sprinter’s average acceleration is 1.98 m/s²
The given parameters;
- initial velocity of the sprinter, u = 18 km/h
- final velocity of the sprinter, v = 27 km/h
- time of motion of the sprinter, t = 3.5 x 10⁻⁴ h
Convert the velocity of the sprinter to m/s;

The time of motion is seconds;

The sprinter’s average acceleration is calculated as follows;

Thus, the sprinter’s average acceleration is 1.98 m/s²
Learn more here:brainly.com/question/17280180
Answer:
35.35 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 20 m/s
Angle of projection (θ) = 30°
Acceleration due to gravity (g) = 9.8 m/s²
Range (R) =.?
The range (i.e how far away) of the ball can be obtained as follow:
R = u² Sine 2θ /g
R = 20² Sine (2×30) / 9.8
R = 400 Sine 60 / 9.8
R = (400 × 0866) / 9.8
R = 346.4 / 9.8
R = 35.35 m
Therefore, the range (i.e how far away) of the ball is 35.35 m