Motion in a Straight Line Test 11

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Motion in a Straight Line Test 11

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Weekly Quiz Competition

Question 1
1 / -0

Speedometer of a car measures

A
average speed

B
average velocity

C
instantaneous speed

D
instantaneous velocity

Solution

Speedometer of the car measures the instantaneous speed of the car.
Question 2
1 / -0

Speedometer measures the speed of the car in

A
m s⁻¹

B
cm s⁻¹

C
km h⁻¹

D
km min⁻¹

Solution

Speedometer measures the speed of the car in km h⁻¹.
Question 3
1 / -0

A particle moves with uniform velocity. Which of the following statements about the motion of the particle is true?

A
Its speed is zero

B
Its acceleration is zero

C
Its acceleration is opposite to the velocity

D
Its speed may be variable

Solution

A particle moving with uniform velocity has zero acceleration.
Question 4
1 / -0

The position of an object moving along x-axis is given by x = a + bt², where a = 8.5 m and b = 2.5 m s⁻² and t is measured in seconds. The instantaneous velocity of the object at t = 2 s

A
5 m s^-1

B
10 m s^-1

C
15 m s^-1

D
20 m s^-1

Solution

Here, x = a + bt²
Where, a = 8.5m and b = 2.5ms⁻²
At t = 2s,
v = 2(2.5ms⁻²)(2s) = 10ms⁻¹
Question 5
1 / -0

The motion of a particle is described by the equation x = a + bt² where a = 15 cm and b = 3 cm/sec². Its instantaneous velocity at time 3 sec will be:

A
33 cm s^-1

B
18 cms^-1

C
16 cm s^-1

D
32 cms ^-1

Solution

here, x = a + bt²
Given a = 15 cm b = 3cm/sec^-2
at t = 3 s, v = 15 + 2 x 3 x 3 = 33cm s^-1
Question 6
1 / -0

The motion of a particle is described by x = x₀(1 − e^−kt); t ≥ 0; x₀ > 0, k > 0. With what velocity does the particle start?

A
x₀/k

B
x₀k

C
k/x₀

D
2x₀k

Solution
Question 7
1 / -0

The velocity of the particle at any time t is given by v = 2t(3 − t)m s⁻¹. At what time is its velocity maximum?

A
2 s

B
3 s

C
2/3s

D
3/2 s

Solution

given v = 2t(3 − t) or v = 6t - 2t²
For V maximum or minimum
Question 8
1 / -0

The motion of a body is given by the equation dv/dt = 6 − 3v where v is the speed in m s⁻¹ and t is time in s. The body is at rest at t = 0. The speed varies with time as

A
v = (1 − e^−3t )

B
v = 2(1 − e^−3t)

C
v = (1 + e^−2t)

D
v = 2(1 + e^−2t)

Solution

Integrating both sides, we get
where C is a constant of integration
At t = 0, v = 0
Question 9
1 / -0

The position x of a particle with respect to time t along x-axis is given by x = 9t² −t³.where x is in metres and t is in seconds. What will be the position of this particle when it achieves maximum speed along the + x direction?

A
54 m

B
81 m

C
24 m

D
32 m

Solution

Given: x = 9t² − t³
for Maximum speed, dv/dt
= 0 ⇒ 18 - 6t = 0
∴ t = 3s.
∴ x_max = 81m - 27m
= 54 m. (From x = 9t² - t³)
Question 10
1 / -0

Figure shows the displacement (x) -time (t) graph of the particle moving on the x-axis.

A
The particle is at rest.

B
The particle is continuously going in positive x-direction.

C
The velocity of the particle increases upto time t₀ and then becomes constant.

D
The particle moves at a constant velocity up to a time t₀and then stops.

Solution

The displacement-time graph is a straight line inclined to time axis upto time t₀ indicates a uniform velocity. After time t₀, the displacement-time graph is a straight line parallel to time axis indicates particle at rest.
Question 11
1 / -0

The figure shows the x−t plot of a particle in one-dimensional motion. Two different equal intervals of time are shown. Let v₁ and v₂be speed in time intervals 1 and 2 respectively. Then

A
v₁ > v₂

B
v₂ > v₁

C
v₁ = v₂

D
Data is insufficient

Solution

The average speed in a small interval of time is equal to the slope of x-t graph during that interval. As slope of interval 1> slope of interval 2
∴ v₁ > v₂
Question 12
1 / -0

The position-time (x - t) graphs for two children A and B returning from their school O to their homes P and Q respectively are as shown in the figure. Choose the incorrect statement regarding these graphs.

A
A lives closer to the school than B

B
A starts from the school earlier than B

C
A walks faster than B

D
A and B reach home at the same time

Solution

As OP < OQ, A lives closer to the school than B.
(b) For x = 0, t = 0 for A; while t has some finite value for B. Therefore,
A starts from the school earlier than B.
(c) Slope of x - t graph is equal to the speed. Since the slope of x-t for B is greater than that for A. Therefore, B walks faster than A.
(d) Corresponding to points P and Q the value of t from x - t graphs for A and B is same. Thus, both A and B reach home at the same time.
Question 13
1 / -0

The displacement-time graph of a moving particle is shown below. The instantaneous velocity of the particle is negative at the point

A
C

B
D

C
E

D
F

Solution

The slope of the tangent at any point on the displacement-time graph gives instantaneous velocity at that instant. In the given graph, the slope is negative at point E.
Question 14
1 / -0

In the given v-t graph the distance traveled by the body in 5 seconds will be

A
100 m

B
80 m

C
40 m

D
20 m

Solution

The distance is equal to total area under v-t graph
Question 15
1 / -0

The speed-time graph of a particle moving along a fixed direction is shown in Fig. Obtain the distance traversed by the particle between time t = 0 s to 10 s.

A
20 m

B
40 m

C
60 m

D
80 m

Solution

Distance = Area under speed-time graph over a given time interval.
∴ Distance traversed by the particle between t = 0 s to t = 10 s will be