1.In one dimensional motion, instantaneous speed v satisfies

0 ≤ v < v0.
(a) The displacement in time T must always take non-negative
values.
(b) The displacement x in time T satisfies – vo T < x < vo T.
(c) The acceleration is always a non-negative number.
(d) The motion has no turning points (choose the correct option and also explain how to solve)

2.The displacement of a particle is given by x = (t – 2)2 where x is in
metres and t in seconds. The distance covered by the particle in
first 4 seconds is
(a) 4 m
(b) 8 m
(c) 12 m
(d) 16 m (choose the correct option and also please explain how to solve)

3.At a metro station, a girl walks up a stationary escalator in time t1.
If she remains stationary on the escalator, then the escalator takeher up in time t2. The time taken by her to walk up on the moving
escalator will be
(a) (t1 + t2)/2
(b) t1t2/(t2–t1)
(c) t1t2/(t2+t1)
(d) t1–t2 (choose the correct option and also explain how to solve)

1.(B)

2.(B)
3.(C)

thanks good one just in time once again thank you for such ugly easy answer you slime

1. The correct option is (d) The motion has no turning points.

To solve this, we need to understand what a turning point is. In one-dimensional motion, a turning point occurs when the velocity changes sign, meaning it goes from positive to negative or vice versa.

In this case, we are given that 0 ≤ v < v0, which means the velocity is always non-negative. Since the velocity can never change sign, there can be no turning points in the motion.

2. The correct option is (d) 16 m.

To find the distance covered by the particle in the first 4 seconds, we need to calculate the displacement at the end of 4 seconds and take the absolute value.

Given x = (t - 2)^2, we substitute t = 4 into the equation:
x = (4 - 2)^2
x = 2^2
x = 4

Since we are interested in the absolute value, the distance covered by the particle in the first 4 seconds is 4 meters.

3. The correct option is (c) t1t2/(t2 + t1).

To solve this problem, we can consider the relative speeds at which the girl is moving and the escalator is moving.

When the girl walks up the stationary escalator, her speed is the sum of her walking speed and the speed of the escalator. Therefore, her speed is (1/t1 + 1/t2).

When the girl remains stationary on the moving escalator, her speed is just the speed of the escalator, which is 1/t2.

So, in the case where the girl is walking up on the moving escalator, her speed will be the difference between the two speeds, which is (1/t1 + 1/t2) - 1/t2.

Now, we can calculate the time taken by the girl to walk up on the moving escalator by dividing the distance traveled by the speed:
Time = Distance/Speed

The distance traveled is the same, so we can set up the equation:
t1 = Distance/(1/t1 + 1/t2 - 1/t2)

Simplifying the equation, we get:
t1 = (t1t2)/(t2 + t1 - t2)
t1 = (t1t2)/(t1)
t1 = t2/(t2 + t1)

Therefore, the correct option is (c) t1t2/(t2 + t1).

1. Since 0 ≤ v < v0 , this is decelerated motion.

(d) The motion has no turning points .

2. x=(t-2)2=2t-4 = 2•4 – 4 =4m
(a) 4 m

3. v(girl)=s/t1
v(esc)=s/t2
t=s/(v1+v2) = t1•t2/(t1+t2)
(c)