Question

A particle Stars moving +ve x direction with initial velocity of 10ms-1 with a uniform acceleration of magnitude 2ms-1 but direcited in - ve x direction. What is the distance travesed by the particle in 12 sec

Answer 1

Answer:

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