Question

consider a particle initially moving with a velocity 5m/s starts retarding at a constant rate of 2 m/s2
Determine the time at which the particle becomes stationery
Find the distance travelled in the second second
Find the distance travelled in the third second

Answer 1

1) As
v=u+at
Final velocity= initial velocity + acceleration * time taken
Therefore
0= 5+ (-2)t
-5= -2t
t= 2.5 sec

2) s=ut + at^2/2
Distance will be
5 x 2 -4
6m

3) you can do by yourself just keeping the value of t as 3 instead of 2

Answer 2

Answer:

The correct answer is using the first equation of the motion.

Explanation:

Given:

Velocity= $5 m/s$

Constant rate = $2m^{2}$

1. Determine the time at which the particle becomes stationery

Answer:

By applying first equation of motion, we get,

$Let v =u +at$

Final velocity= initial velcoty+acceleration* time taken

$0 = 5+ (-2) * t-5 = -2t\\T= 2.5 seconds$

2. Find the distance travelled in the second second:

Answer:

$Let s = ut+at^ 2 /2$

The distance in traveled in second will be:

$5 * 2 (-4)$

$=10-4=6 m$

3. Find the distance travelled in the third second

Answer:

$s = 5 *3(-4)$

$s= 15+4s= 20 m$

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