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:

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.

Answer 2

Answer: $857\ m$

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

$\Rightarrow s_d=30\times t$

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

$s_t=ut+\dfrac{1}{2}at^2\\\\s_t=0+\dfrac{1}{2}\times 2.1t^2$

now, $s_d=s_t$

$30t=\dfrac{2.1}{2}t^2\\\\\Rightarrow 2.1t^2-60t=0\\\Rightarrow t(2.1t-60)=0\\\Rightarrow t=0,28.57\ s$

Neglecting $t=0$

Distance traveled by tina in $28.57\ s$ is

$s_t=\dfrac{1}{2}\times 2.1\times (28.57)^2\\\\s_t=857.057\approx 857\ m$