David is driving a steady 28.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.90 m/s^2 at the instant when David passes. Required:
a. How far does Tina drive before passing David?
b. What is her speed as she passes him?

Explanation:

Let t represent the time for Tina to catch David.

Hence, considering the equation of linear motion S = ut + 1/2at^2 1

For David u = 28.0 m/s where 'a' is set to nought

S = ut

S = 28t2

For Tina consider equation 1

Where acceleration = 2.90m/s^2 and u is set at nought

S = 1/2×2.90 m/s×t^23

Equate 2 and 3

28t = 1.45t^2

Divide through by t

28 = 1.45t

t = 28/1.45

t = 19.31seconds

Now put the value of t into equation 3

S = 1/2×2.90 m/s×t^23

= 1.45×20×20

= 580m

Tina must have driven 580meters before passing David

Considering the equation of linear motion : V^2 = U^2+2as

Where u is set at nought

V^2 = 2as

V^2 = 2×2.9×580

V^2 = 3364

V = √3364

V = 58m/s

Her speed will be 58m/s

it was given as a lesson and guide for prayer.