A particle moving on the x axis with constant acceleration has displacement of 6m from t=4sec to t=7sec and 3m from t=5sec to t=8sec. The distance covered fromt=6sec to t=9sec is?
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Answered by
17
s = ut + 1/2 at²
6 = u(7-4) + 1/2a(7² - 4²)
6 = 3u + 33/2a
Similarly
3 = 3u + 39/2a
by solving equations we get
u = 7.5 m/s
and a = -1 m/s²
we know s = ut + 1/2at²
s = 7.5 x (9-6) + 1/2(-1) (9² - 6²)
s = 22.5 - 22.5 = 0 m
6 = u(7-4) + 1/2a(7² - 4²)
6 = 3u + 33/2a
Similarly
3 = 3u + 39/2a
by solving equations we get
u = 7.5 m/s
and a = -1 m/s²
we know s = ut + 1/2at²
s = 7.5 x (9-6) + 1/2(-1) (9² - 6²)
s = 22.5 - 22.5 = 0 m
Answered by
20
Answer: 2.25m
Explanation:since it is clear from the given data that velocity is reversed somewhere in between so we assume that at X sec velocity reverses.
Finally solving the 2eqations as shown in attachment we get value of X as 7.5 sec. Hence to calculate the distance travelled by particle from 6 to 9 sec we break it in two segments and calculate as shown.
Attachments:
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