Question

2) A body is moving with an initial velocity of 2im/s and acceleration of 0.5m
time when the velocity of the particle becomes zero.
​

Answer 1

ANSWER
a. 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 x
n
​
=u+
2
a
​
(2n−1)=5−
2
2
​
[2(2)−1)]=2m
c. Here, if we use the above formula, we will get x
n
​
= 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
x
2.5
​
=
2a
u
2

​
=
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.,
x
3
​
=x
2
​
=5(2)+(1/2)(−2)(2)
2
= 6m
Thus, the distance travelled in the third second is
x=(x
2.5
​
−x
2
​
)+(x
2.5
​
−x
3
​
)
= ( 6.25 - 6 ) + ( 6.25 - 6 ) = 0.5m






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