Answer
(a) The horizontal acceleration was $a = 1.0 \times 10^1 ~m/s^2$
(b) The sprinter left the starting block with a speed of 3.3 m/s.
Work Step by Step
(a) $F_x = F ~cos(\theta) = 720 ~cos(22^{\circ}) ~N = 670 ~N$
We can use the horizontal force to find the acceleration.
$ma = F_x$
$a = \frac{F_x}{m} = \frac{670 ~N}{65 ~kg} = 1.0 \times 10^1~m/s^2$
The horizontal acceleration was $a = 1.0 \times 10^1 ~m/s^2$
(b) $v = v_0 +at = 0 + (10.3 ~m/s^2)(0.32 ~s) = 3.3 ~m/s$
The sprinter left the starting block with a speed of 3.3 m/s.
