Question

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?

Answer 1

Answer:

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 = 28t.......2

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^2.......3

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^2.......3

= 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

Answer 2

(a) Tina should drive for 580 m, before passing the David.

(b) The speed of Tina during her passage through the David is 58 m/s.

Given data:

The initial velocity of the David is, u = 28.0 m/s.

The magnitude of acceleration is, $a = 2.90 \;\rm m/s^{2}$.

(a)

We can use the second kinematic equations of motion to obtain the distance covered by Tina, before passing the David. As per the second kinematic equation of motion,

$s= u't + \dfrac{1}{2}at^{2}$

Here, u' is the initial speed of Tina and t is the time interval. Then,

Let t represent the time for Tina to catch David.

Hence, considering the equation of linear motion as,

S = ut + 1/2at²...............................................................(1)

Also,

S = ut

S = 28t ...........................................................................(2)

For Tina consider equation 1

S = 1/2×2.90t²................................................................(3)

Equate 2 and 3

28t = 1.45t²

 28 = 1.45t

t = 28/1.45

t = 19.31 seconds

Now put the value of t into equation (3)

S = 1/2×2.90 t².

   = 1.45×20×20

   = 580m

Thus, we can conclude that Tina should drive for 580 m, before passing the David.

(b)

Now, using the third kinematic equation of motion to obtain the speed of Tina during her passage through David as,

v² = u²+2as

Solving as,

v² = 28.0² + 2(2.90)(580)

v = √3364

v = 58m/s

Thus, we can conclude that the speed of Tina during her passage through the David is 58 m/s.

Learn more about the Kinematic equation of motion here:

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