0
$\begingroup$

A particle starts from rest and moves in a straight line with constant acceleration. In a certain 4 seconds of its motion it travels 12 m and in the next 5 seconds it travels 30m.

The acceleration of the particle is? The velocity of the particle at the start of the timing is? The distance it had travelled before timing started was?

Apparently, the velocity of the particle at the start of the timing is not 0 m/s

$\endgroup$
  • 2
    $\begingroup$ At the midpoint (on the time-axis) of the first interval the speed was 3m/s. At the midpoint of the latter interval the speed was 6m/s. So... $\endgroup$ – Jyrki Lahtonen Sep 25 '11 at 14:27
1
$\begingroup$

Use $s=ut+\frac{1}{2}at^2$.We are told that $u=0$.
Let the timing start at time $t$ and at distance $s$.

$s=\frac{1}{2}at^2$
$s+12=\frac{1}{2}a(t+4)^2$
$s+42=\frac{1}{2}a(t+9)^2$

Solving these gives $a=\frac{2}{3}$, $t=\frac{5}{2}$ and $s=\frac{25}{12}$.

For the velocity, use $v^2=u^2+2as$.

$\endgroup$
0
$\begingroup$

Let's observe picture bellow.We can write next equations:

$x=\frac{1}{2}at_x^{2}$ ; $v_A=at_x$; $v_B=v_A+at_1$; $s_1=v_At_1+\frac{1}{2}at_1^{2}$; $s_2=v_Bt_2+\frac{1}{2}at_2^{2}$

$s_1=4at_x+8a \Rightarrow 12=4at_x+8a \Rightarrow at_x=3-2a$

$s_2=(at_x+4a)5+\frac{25}{2}a \Rightarrow 6=at_x+4a+\frac{5}{2}a \Rightarrow 12=2at_x+13a$ , Since that :

$12=2(3-2a)+13a \Rightarrow a=\frac{2}{3}\frac{m}{s^2} \Rightarrow t_x=\frac{5}{2}sec$, $ v_A=at_x \Rightarrow v_A=\frac{5}{3} \frac{m}{s}$ and $x=\frac{1}{2}at_x^{2} \Rightarrow x=\frac{25}{12} m$

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.