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# phy question * so important ,, before 12pm 20/10* +20marks

an object, beginning at x=0, is traveling along the x-axis for 15 seconds at a velocity of

+6 m/s . at t=0, an acceleration of -13m/s^2 is imposed upon it.At what point in time does

the object return to x=0?

### 2 Answers

- Anonymous1 decade agoFavorite Answer
D = Vt

15 second travel distance = 15 * 6 = 90m

At x = 90 m, deceleration applied.

a = V/t, => 13 = 6/t => t = 6/13s to stop the object.

D = Vt => D = at^2 => D = 13 * (6/13)^2 = 36/13 m before it stop.

total distance from x = 0 is 36/13 + 90

t to return to 0 is 90 + 36/13 = 13t ^2

(90 + 36/13)/13 = t^2

t = 2.67 s

Source(s): CLOUDWAH - pLv 61 decade ago
Note: The previous approach is correct in general, but in the formula

S=ut+(1/2)at^2 where the (1/2) was missing. As a result, the answer is not correct.

For this particular question, it is not necessary to determine the stationary position.

Question:

an object, beginning at x=0, is traveling along the x-axis for 15 seconds at a velocity of +6 m/s . at t=0, an acceleration of -13m/s^2 is imposed upon it.

At what point in time does the object return to x=0?

The object moved 6*15=90m after 15 seconds at 6m/s.

Denote time=t=0 at this instant, and the distance S=0, velocity=u=6 m/s

to go back to the starting point, S=-90.

At time t, the general formula for distance

S(t)=ut+(1/2)at^2

For, S=-90,

-90=6t+(-13)t^2/2

from which

t=-(6*sqrt(66)-6)/13 or t=(6*sqrt(66)+6)/13

We are not interested in the solution where t<0, so

t=(6*sqrt(66)+6)/13

=4.21109 sec. AFTER the 15 seconds at uniform speed of 6 m/s.