Question

A particle moving along a straight line with a constant acceleration of -4 m/ s^2 passes through a point A on the on the line with a velocity of +8 m/ s at some moment . find the distance travelled by the particle in 5 second after that moment.?

Answer 1

HERE is u r ANSWER



u = 8 m/ s


a= -4m/s^2


v= 0

= > 0 = 8 - 4t. or t = 2sec


displacement in first 2 sec.



S1 =
$\: 8 \times \frac{1}{2} \times ( - 4) {2}^{2} \: = 8 \\ m$


displacement in next 3sec.



S2 =

$0 \times 3 + \frac{1}{2} ( - 4) {3}^{2} = \: = - 18m$


distance travelled =

$|s1| + |s2| = 26m$



Ans. :- 26m

Answer 2

ANSWER OF YOUR QUESTION-:

u = 8 m/ s 


a= -4m/s^2


v= 0

= > 0 = 8 - 4t. or t = 2sec


displacement in first 2 sec.



S1 = 
\begin{lgathered}\: 8 \times \frac{1}{2} \times ( - 4) {2}^{2} \: = 8 \\ m\end{lgathered}8×21​×(−4)22=8m​ 


displacement in next 3sec.


S^2 =0 \times 3 + \frac{1}{2} ( - 4) {3}^{2} = \: = - 18m0×3+21​(−4)32==−18m 
distance travelled =|s1| + |s2| =26m∣s1∣+∣s2∣ =26m 

=26m 《《ANSWER》》