Answer
(a) Magnitude: $2.1~m/s^2$, direction: East.
(b) $0.97~s$
(c) Less than 3.4 m/s.
Work Step by Step
(a) We know that
$a=\frac{v_f^2-v_i^2}{2\Delta x}$
We plug in the known values to obtain:
$a=\frac{(6.4m/s)^2-(+8.4m/s)^2}{2(7.2m)}$
$a=-2.1m/s^2$
The negative sign shows that its direction is towards east.
(b) We can find the required time as
$t=\frac{v_f-v_i}{a}$
We plug in the known values to obtain:
$t=\frac{(6.4m/s)-(8.4m/s)}{-2.1m/s^2}$
$t=0.97s$
(c) We know that
$v_f^2=v_i^2+2a(\Delta x)$
We plug in the known values to obtain:
$v_f^2=(5.4m/s)^2+2(-2.055m/s^2)(7.2m)$
This simplifies to:
$v_f=-0.432m/s$
This shows that the bicycle comes to a halt without crossing the sandy patch and hence the final speed is less than $3.4m/s$.
