A particle moves in a straight line, starting from rest at point $P$. It accelerates for $5$ seconds, until it reaches a speed of $16\frac m s$. It maintains this speed for $T$ seconds and then decelerates at $2 \frac{m}{s^2}$ until it comes to a rest at point $Q$. Given that the average speed of the particle on the journey from $P$ to $Q$ is $12 \frac m s$, find the value of $T$. How would I do this without the distance? I’ve found the time it took to come back to rest, but not very sure where to go from there and what to do with the given average speed.
Asked
Active
Viewed 59 times
-1
-
Represent the distance travelled in terms of $T$ in two ways, one by adding the distances of the 3 stages, and one by the given average speed. – peterwhy Feb 07 '21 at 17:29
1 Answers
0
The distance travelled till the speed is $16$ can be obtained using the relation $v^2-u^2=2as$:
$$ 16^2 -0^2 =2 a s =2\frac{16-0}{5} s =16s \implies s=40$$ where I have further used $a=\frac{v-u}{t} $.
The distance travelled during the $T$ seconds is $16T$, and that while decelerating can be calculated by using $v^2-u^2 =2as$:
$$0^2-16^2 = 2(-2) s \implies s=64 $$ Now that you have the total distance and total time (and the average speed), you can set up a linear equation in $T$.
Vishu
- 14,469