Question

A particle starts moving in + ve x direction with initial velocity of 10 ms–1 with a uniform acceleration of

magnitude 2 ms−2

but directed in – ve x direction. What is the distance traversed by the particle in 12 seconds :

Answer 1

Final Answer: 74m 

Steps:
1) Since,velocity and acceleration are opposite .
=> Velocity will become 0 and change its direction at some time .
 Initial velocity, u = 10im/s  
acceraltion ,a = -2i $m/s^{2}$ 
Final velocity, v= 0m/s
Then,we
v=u+at 
=> 0=10-2t 
=> t=5s 

2) Distance covered from t=0 to t=5s,is same as Displacement .

Displacement ,
$s=ut + \frac{1}{2}at^{2}$ 
 [tex]S= 10*5 - \frac{1}{2}*2* 5^{2} \\ S = 50 -25 = 25 m [/tex] 
Hence,Distance covered from t=0 to t=5s is 25m.

3) Now,direction of velocity is reversed and initial velocity at t=5s is 0. 
Motion from t= 5 to t=12s .
Change in t = 7s .

[tex]S=0*7 - \frac{1}{2}*2*7^{2} = -49m \\ \\ =\ \textgreater \ x-25 =-49m \\ \\ =\ \textgreater \ x= -24m. [/tex]

Hence ,Distance covered from t=5s to t=12 s is (25-(-24)) =49m. 

Therefore,Total Distance covered from t=0s to t=12s is
$\boxed{49+25=74m}$ 

Answer 2

Now, the particle initially starts with velocity 10m/s
But acceleration is in negative x direction,
So, there will be certain time taken to stop the particle and then transverse back in negative x direction
a=- 2  m/s²
Now,
v=u+ at
v= 10  + (-2)t
v=0
t=5 sec
So, out of 12 sec , 5 sec will be taken for stopping the particle
Now, for remaining 7 sec distance covered in -ve x direction will be,
s= ut + 1/2at²
s= 1/2 ×  2 × 7²
s=49m
So,
It will travel 49 m in negative x direction
For first 5 sec distance travelled will be,
s= ut+ (1/2)at²
s=10× 5 -1/2 × 2 ×25
s=50-25
s=25m

So,
Total distance covered will be 49+25= 74 m