Question

the body is moving with an initial velocity of 2m/s and acceleration of -0.5m/s².Find the time when the velocity of the particle becomes zero​

Answer 1

Explanation:

here according to the question:

initial velocity (u) =2m/s

acceleration=(-0.5)m/s^2

final velocity(v) =0m/s

let the time be=t

we know that v=u+at (first equation of motion)

so 0=2 + (-0.5×t)

-2=(-1/2t)

so, t=4

therefore the time taken is 4 seconds

Answer 2

Answer:

Here u = 5ms−1, a = -2 ms−2, v= 0, t = ?

Using v = u + at, where

0 = 5 - 2t ⇒ t = 2.5s

b. Here u = 5ms−1, a = -2 ms−2, n = 2.

Using xn=u+2a(2n−1)=5−22[2(2)−1)]=2m

c. Here, if we use the above formula, we will get xn = 0, but in reality, it is not zero. This formula is not applicable for the third second because velocity becomes zero in the third second, i.e., at t = 2.5 s. The particle has a turning point at t = 2.5 s. We have to indirectly calculate the distance travelled in this particular second. That is, we have to determine the distance travelled, between 2 ≤ t ≤ 2.5 and 2.5≤t≤3, and then add the two.

Displacement of the particle at t = 2.5 s is

x2.5=2au2=2(2)(5)2 = 6.25m

Due to symmetry, the displacement of the particle at t = 2 s and t = 3 s are same, i.e.,

x3=x2=5(2)+(1/2)(−2)(2)2 = 6m

Thus, the distance travelled in the third second is

x=(x2.5−x2)+(x2.5−x3)

= ( 6.25 - 6 ) + ( 6.25 - 6 anics