A sprinter accelerates from rest at a uniform rate of 1.75m/s2 for 4 sec to reach his maximum speed.?

a.
Initial velocity, u = 0 m/sed
Acceleration, a = 1.75 m/sec²
Time taken, t = 4 sec

s = u t + (1/2) a t²
s = 0 + (1/2) × 1.75 × 4²
Distance traveled, s = 14 m

====
b.
Initial velocity, u = 0 m/sec
Acceleration, a = 1.75 m/sec²
Time taken, t = 4 sec

v = u + a t
v = 0 + 1.75 × 4
Speed after 4 sec, v = 7 m/sec

====
c.
Initial velocity, u = 0 m/sec
Acceleration, a = 1.75 m/sec²
Displacement, s = 12 m

s = u t + (1/2) a t²
12 = 0 + (1/2) × 1.75 × t²
t = √(12 × 2 / 1.75)
Time taken, t = 3.7 sec

====
d.
Initial velocity, u = 0 m/sec
Acceleration, a = 1.75 m/sec²
Displacement, s = 12 m

v² = u² + 2 a s
v² = 0 + 2 × 1.75 × 12
v = √(2 × 1.75 × 12)
Speed when passing the 12 m point, v = 6.48 m/sec

You need the basic acceleration and distance formulas. Plug in the values and solve.
Most come from: x(t) = ½at² + v₀t  + x₀

u = 0, a = 1.75 and t = 4

Using s = ut + (1/2)at² we get:

s = (1/2)(1.75)(4)² => 14 metres

Using v = u + at we get:

v = 1.75(4) => 7 ms⁻¹

Again, using s = ut + (1/2)at² we get:

12 = (1/2)(1.75)t²

so, t² = 24/1.75

Then, t = 3.7 seconds

Again, using v = u + at we get:

v = 1.75(3.7) => 6.48 ms⁻¹

:)>

a(t) = 1.75
v(t) = 1.75 * t + C
v(0) = 0
v(t) = 1.75 * t
s(t) = (1.75 / 2) * t^2 + C
s(0) = 0
s(t) = 0.875 * t^2


a(t) = 1.75
v(t) = 1.75 * t
s(t) = 0.875 * t^2

a)
t = 4
s(4) = 0.875 * 4^2 = (7/8) * 16 = 14 meters

b)
v(4) = 1.75 * 4 = 7 m/s

c)
s(t) = 12
12 = (7/8) * t^2
12 * (8/7) = t^2
4 * 4 * 2 * 3 / 7 = t^2
16 * 6 * 7 / 49 = t^2
(4/7) * sqrt(42) = t
t = (4/7) * sqrt(42) seconds

d)
t = (4/7) * sqrt(42)
v(t) = (7/4) * (4/7) * sqrt(42) = sqrt(42) meters

For the last 2, take it to the appropriate significant figures.  I'd go with 3.