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v = 40 - 10t
₀

^x dx = ₀

^t (40 - 10t) dt
x = 40t - 5t²
Now at t = 4 s, the particle comes to rest. That is it turns around towards the -ve X axis at t = 4 s.
Now, position of the particle at t = 4 s is
x(4 s) = 80 m.
Hence, the particle once crosses the x = +60m point before turning, and then again when its retuning back. Moreover, when it crosses the origin to go into the negative X axis, it reaches the point x = -60 m at some instant and then continues to move forward.
Hence there are three time instants when the particle is at a distance of 60 m from the origin, twice at +60m and once at -60m.
For x = +60 m,
40t - 5t² = 60
t = 2 s, 6 s
For x = -60 m,
40t - 5t² = -60
t = 4 + 2

7 s (discarding the -ve root)

Hence the time instants when the particle is at a distance of 60 m from the origin are
t = 2 s, 6 s, 4 + 2

7 s

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