Question

David is driving a steady 30.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.10 m/s2 at the instant when David passes. Part A How far does Tina drive before passing David

Answer 1

Answer:
`857\ m`

Step-by-step explanation:

Given

Speed of David car
`v=30\ m/s`

Tina begins to accelerate
`2.1\ m/s^2` after David pass the tina

Suppose it took t time for tina to catch David

Distance traveled by David in t time

`\Rightarrow s_d=30* t`

Using the equation of motion to get the distance of Tina is

`s_t=ut+(1)/(2)at^2\\\\s_t=0+(1)/(2)* 2.1t^2`

now,
`s_d=s_t`

`30t=(2.1)/(2)t^2\\\\\Rightarrow 2.1t^2-60t=0\\\Rightarrow t(2.1t-60)=0\\\Rightarrow t=0,28.57\ s`

Neglecting
`t=0`

Distance traveled by tina in
`28.57\ s` is

`s_t=(1)/(2)* 2.1* (28.57)^2\\\\s_t=857.057\approx 857\ m`

Answer 2

Step-by-step explanation:

Let after time t , Tina catches up David .

Distance travelled by them are equal ,

Distance travelled by Tina

s = ut + 1/2 a t²

= .5 x 2.10 t²

= 1.05 t²

Distance travelled by David

= 30 t ( because of uniform velocity )

1.05 t² = 30t

t = 28.57 s

Distance travelled by Tina

= 1/2 a t²

= .5 x 2.10 x 28.57²

= 857 m approx.