91Ó°ÊÓ

Deceleration is the rate at which velocity reduces with respect to time.

Initial velocity of Sally,$v_{\mathrm{s}}=5{\mathrm{ms}}^{-1}$

Initial velocity of Mary,$v_{\mathrm{m}}=4{\mathrm{ms}}^{-1}$

Acceleration of Sally,$a_{\mathrm{s}}=-0.4{\mathrm{ms}}^{-2}$

Let the acceleration of Mary be a.

Sally is (22–5) m from the finish line.

Applying the second equation of motion for the remaining part of the race for Sally,

$22-5=5×t+\frac{1}{2}×\left(-0.4\right)×t^2$

The above equation can be rewritten as.

$0.2t^2-5t+17=0$

Solving the above quadratic equation, you get.

t = 20.94 s

or

t = 4.06 s

Take the smaller value as the greater value represents the time when Sally keeps deaccelerating and return to the finishing line.

Thus, the time taken by Sally to complete the remaining part of the race is 4.06 s.

For Mary and Sally to finish the race simultaneously, the time taken by Mary to complete the remainder 22 m of the race is4.06 s.

From the second equation of motion,

$22=v_{\mathrm{m}}t+\frac{1}{2}a{t}^2$

Substituting the values in the above equation,

role="math" localid="1642848891327" $22=\left(4×4.06\right)+\frac{1}{2}a×{\left(4.06\right)}^2$

Solving for a,

$a=0.70{\mathrm{ms}}^{-2}$

Thus, the acceleration of Mary should be $0.70{\mathrm{ms}}^{-2}$.