Question

A jogger accelerates at a constant rate as she travels 5.0
meters along a straight track from point A to point B ,as
shown in the diagram below.
If her speed was 2.0 meters per second at point A and
will be 3.0 meters per second at point B, how long will it
take her to go from A to B?
A) 1.0 s B) 2.0 s C) 3.3 s D) 4.2 s

Answer 1

Answer:

2.0s

Explanation:

...

Answer 2

The time taken by the jogger to jog from point A to Point B is 2 seconds.

Option B) 2.0s is the correct answer.

Given the data in the question;

• Distance from point A to point B; $s = 5.0m$

• Initial velocity or speed; $u = 2.0m/s$

• Final velocity; $v = 3.0m/s$

• Time taken to reach B from A; $t = \ ?$

First we find the acceleration of the jogger, using the Third Equation of Motion:

$v^2 = u^2 + 2as$

Where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance travelled.

So we substitute our values into the equation

$(3.0m/s)^2 = ( 2.0m/s)^2 \ +\ ( 2\ *\ a\ *\ 5.0m)\\\\9m^2/s^2 = 4m^2/s^2 \ +\ (10m \ * \ a)\\\\a = \frac{9m^2/s^2\ -\ 4m^2/s^2}{10m} \\\\a = \frac{5m^2/s^2}{10m} \\\\a = 0.5m/s^2$

Now, we calculate the time it took the jogger to reach B from A.

From the First Equation of Motion:

$v = u + at$

Where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

We make time "t", the subject of the formula

$t = \frac{v -u}{a}$

We substitute our values into the equation

$t = \frac{3.0m/s\ - \ 2.0m/s}{0.5m/s^2} \\\\t = \frac{1.0m/s}{0.5m/s^2}\\\\t = 2s$

Therefore, the time taken by the jogger to jog from point A to Point B is 2 seconds.

Option B) 2.0s is the correct answer.

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