Question

David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.0 m/s2 at the instant when David passes. How far does Tina drive before passing David

Answer 1

Answer:

The distance traveled by Tina before passing David is 900 m

Given:

Initial speed of David, $u_{D} = 30 m/s$

Acceleration of Tina, $a_{T} = 2.0 m/s^{2}$

Solution:

Now, as per the question, we use 2nd eqn of motion for the position of David after time t:

$s = u_{D}t + \frac{1}{2}at^{2}$

where

s = distance covered by David after time 't'

a = acceleration of David = 0

Thus

$s = 30t$

Now, Tina's position, s' after time 't':

$s' = u_{T}t + \frac{1}{2}a_{T}t^{2}$

where

$u_{T} = 0$, initially at rest

$s' = 0.t + \frac{1}{2}\times 2t^{2}$

$s' = t^{2}$                     (1)

At the instant, when Tina passes David, their distances are same, thus:

s = s'

$30t = t^{2}$

$t(t - 30) = 0$

t = 30 s

Now,

The distance covered by Tina before she passes David can be calculated by substituting the value t = 30 s in eqn (1):

$s' = 30^{2}$ = 900 m