Motion in A Plane
Here, we have
Initial velocity, u = 200 cm/s
Acceleration, a = -10 cm/s2
Distance travelled, s = 1500 cm
Time taken, t = ?
Using the relation,
s = ut + 1/2 at2
1500 = 200 t + 1/2(-10)t2
On solving the equation,we get
t = 10 s or t = 30 s
Here, the value of 10 s corresponds to the time when the particle first arrives at the given location. It then crosses this point and at the end of 20 s, its velocity is just 0.
The particle then returns and at the end of 10 s more, is again at the given location from the starting point.
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