Question

0.6. A particle accelerates from rest at a constant rate for sometime and attains a constant velocity of 8 ms-1. Afterwards it decelerates with a constant rate and comes to rest. If the total time taken is 4 second, the distance travelled is : (a) 32 metre (b) 16 metre (d) insufficient data (c) 4 metre​

Answer 1

Answer:

16m

Solution:-

Time taken to attain a velocity 'v' from rest

                                       = Time taken to come at rest from velocity 'v'

Total time = 4sec

So, Time from rest to 'v' = from 'v' to rest = 4/2 = 2sec

Now,

For t=0 to t=2 sec, we find acceleration, by

v = u + at

8 = 0+ a(2)

a = 8/2 = 4 $m/s^{2}$

From, Equation of motion for distance 's'

$s= ut+ \frac{1}{2} at^{2}$

$s= 0t + \frac{1}{2} (4)( 2^{2})$

$s= \frac{1}{2} (4) (4)$

s = 8m

Now,

As it is a case of constant acceleration from t=0 to t=2 sec & from t=2 to t=4

∴ Distance travelled during t=0 to 2 sec = Distance travelled during t=2 to 4 sec = s

So total distance = 2s = 16m

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