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PHXI03:MOTION IN A STRAIGHT LINE

362315 The speed-time graph of a particle moving along a solid curve is shown below. The distance traversed by the particle from $t = 0 s$ to $t = 3 s$ is
supporting img

1

5 m

2

\frac{5}{2} m

3

\frac{10}{3} m

4

2 m

Explanation:

supporting img
Distance travelled by the particle ,
$s =$ Area under the curve
$= s_{01} + s_{12} + s_{23}$
$s =$ Area of $△ O A B$
+ Area of trapezium A B C D
$+ A r e a o f Δ D C E$
$= \frac{1}{2} \times O B \times A B + \frac{1}{2} (A B + C D) \times B D + \frac{1}{2}$
$\times D E \times D C$
$= \frac{1}{2} \times (1 - 0) \times 1 + \frac{1}{2} \times (1 + 1.5) \times (2 - 1)$
$+ \frac{1}{2} \times (3 - 2) \times 1.5$
$= \frac{1}{2} + \frac{1}{2} \times 2.5 + \frac{1}{2} \times 1.5$
[ $∵$ area of triangle $= \frac{1}{2} \times$ base $\times$ height]
$a r e a o f t r a p e z i u m = \frac{1}{2} \times$
$(s u m o f p a r a l l e l s i d e s) \times h e i g h t]$
$= \frac{1}{2} (1 + 2.5 + 1.5) = \frac{1}{2} \times 5 = \frac{5}{2} m$

PHXI03:MOTION IN A STRAIGHT LINE

362316 A body travels $102.5 m$ in $n^{th}$ second and $115.0 m$ in $(n + 2)^{th}$ second. The acceleration is

1

9 m / s^{2}

2

6.25 m / s^{2}

3

5 m / s^{2}

4

12.5 m / s^{2}

Explanation:

Distance travelled in $n^{th}$ second is
Use $S_{n^{t h}} = u + \frac{1}{2} a (2 n - 1)$
$102.5 = u + \frac{1}{2} a (2 n - 1)$ $(1)$
$115 = u + \frac{1}{2} a (2 \times (n + 2) - 1)$
$115 = u + \frac{1}{2} a (2 n + 3) (2)$
Subtract equation (1) from (2), we get
$115 - 102.5 = \frac{1}{2} a (2 n + 3 - 2 n + 1)$
$\Rightarrow a = 6.25 m / s^{2}$

JEE - 2024

PHXI03:MOTION IN A STRAIGHT LINE

362317 An electron starting from rest has a velocity that increases linearly with time as $v = k t$ , where $k = 2 m / s^{2}$ . The distance covered in the first
3 seconds will be :

1 9

m

2 16

m

3 27

m

4 36

m

Explanation:

Variation of velocity with $t$ is given as $v = k t$
$\Rightarrow \frac{d x}{d t} = k t$
$\Rightarrow d x = k t d t$
Distance covered in first 3 seconds will be
$\int_{0}^{x} d x = k \int_{0}^{3} t d t$
$\Rightarrow x = \frac{1}{2} k {[t^{2}]}_{0}^{3} = \frac{1}{2} \times 2 \times 9$
$∴ x = 9 m$ .
So correct option is (1)

PHXI03:MOTION IN A STRAIGHT LINE

362318 A car starting from rest accelerates at the rate $f$ through a distance $s$ , then continues at a constant speed for time $t$ and then decelerates at the rate $f / 2$ to come to rest. If the total time taken is $15 s$ , then

1

s = f t

2

s = \frac{1}{6} f t^{2}

3

s = \frac{1}{72} f t^{2}

4

s = \frac{1}{4} f t^{2}

Explanation:

supporting img
$v = \sqrt{2 f s} \Rightarrow 12 s = \sqrt{2 f s t}$
$\Rightarrow 144 s^{2} = 2 f s t^{2}$
$\Rightarrow s = \frac{1}{72} f t^{2}$

AIIMS - 2008

PHXI03:MOTION IN A STRAIGHT LINE

362315 The speed-time graph of a particle moving along a solid curve is shown below. The distance traversed by the particle from $t = 0 s$ to $t = 3 s$ is
supporting img

1

5 m

2

\frac{5}{2} m

3

\frac{10}{3} m

4

2 m

Explanation:

supporting img
Distance travelled by the particle ,
$s =$ Area under the curve
$= s_{01} + s_{12} + s_{23}$
$s =$ Area of $△ O A B$
+ Area of trapezium A B C D
$+ A r e a o f Δ D C E$
$= \frac{1}{2} \times O B \times A B + \frac{1}{2} (A B + C D) \times B D + \frac{1}{2}$
$\times D E \times D C$
$= \frac{1}{2} \times (1 - 0) \times 1 + \frac{1}{2} \times (1 + 1.5) \times (2 - 1)$
$+ \frac{1}{2} \times (3 - 2) \times 1.5$
$= \frac{1}{2} + \frac{1}{2} \times 2.5 + \frac{1}{2} \times 1.5$
[ $∵$ area of triangle $= \frac{1}{2} \times$ base $\times$ height]
$a r e a o f t r a p e z i u m = \frac{1}{2} \times$
$(s u m o f p a r a l l e l s i d e s) \times h e i g h t]$
$= \frac{1}{2} (1 + 2.5 + 1.5) = \frac{1}{2} \times 5 = \frac{5}{2} m$

PHXI03:MOTION IN A STRAIGHT LINE

362316 A body travels $102.5 m$ in $n^{th}$ second and $115.0 m$ in $(n + 2)^{th}$ second. The acceleration is

1

9 m / s^{2}

2

6.25 m / s^{2}

3

5 m / s^{2}

4

12.5 m / s^{2}

Explanation:

Distance travelled in $n^{th}$ second is
Use $S_{n^{t h}} = u + \frac{1}{2} a (2 n - 1)$
$102.5 = u + \frac{1}{2} a (2 n - 1)$ $(1)$
$115 = u + \frac{1}{2} a (2 \times (n + 2) - 1)$
$115 = u + \frac{1}{2} a (2 n + 3) (2)$
Subtract equation (1) from (2), we get
$115 - 102.5 = \frac{1}{2} a (2 n + 3 - 2 n + 1)$
$\Rightarrow a = 6.25 m / s^{2}$

JEE - 2024

PHXI03:MOTION IN A STRAIGHT LINE

362317 An electron starting from rest has a velocity that increases linearly with time as $v = k t$ , where $k = 2 m / s^{2}$ . The distance covered in the first
3 seconds will be :

1 9

m

2 16

m

3 27

m

4 36

m

Explanation:

Variation of velocity with $t$ is given as $v = k t$
$\Rightarrow \frac{d x}{d t} = k t$
$\Rightarrow d x = k t d t$
Distance covered in first 3 seconds will be
$\int_{0}^{x} d x = k \int_{0}^{3} t d t$
$\Rightarrow x = \frac{1}{2} k {[t^{2}]}_{0}^{3} = \frac{1}{2} \times 2 \times 9$
$∴ x = 9 m$ .
So correct option is (1)

PHXI03:MOTION IN A STRAIGHT LINE

362318 A car starting from rest accelerates at the rate $f$ through a distance $s$ , then continues at a constant speed for time $t$ and then decelerates at the rate $f / 2$ to come to rest. If the total time taken is $15 s$ , then

1

s = f t

2

s = \frac{1}{6} f t^{2}

3

s = \frac{1}{72} f t^{2}

4

s = \frac{1}{4} f t^{2}

Explanation:

supporting img
$v = \sqrt{2 f s} \Rightarrow 12 s = \sqrt{2 f s t}$
$\Rightarrow 144 s^{2} = 2 f s t^{2}$
$\Rightarrow s = \frac{1}{72} f t^{2}$

AIIMS - 2008

PHXI03:MOTION IN A STRAIGHT LINE

362315 The speed-time graph of a particle moving along a solid curve is shown below. The distance traversed by the particle from $t = 0 s$ to $t = 3 s$ is
supporting img

1

5 m

2

\frac{5}{2} m

3

\frac{10}{3} m

4

2 m

Explanation:

supporting img
Distance travelled by the particle ,
$s =$ Area under the curve
$= s_{01} + s_{12} + s_{23}$
$s =$ Area of $△ O A B$
+ Area of trapezium A B C D
$+ A r e a o f Δ D C E$
$= \frac{1}{2} \times O B \times A B + \frac{1}{2} (A B + C D) \times B D + \frac{1}{2}$
$\times D E \times D C$
$= \frac{1}{2} \times (1 - 0) \times 1 + \frac{1}{2} \times (1 + 1.5) \times (2 - 1)$
$+ \frac{1}{2} \times (3 - 2) \times 1.5$
$= \frac{1}{2} + \frac{1}{2} \times 2.5 + \frac{1}{2} \times 1.5$
[ $∵$ area of triangle $= \frac{1}{2} \times$ base $\times$ height]
$a r e a o f t r a p e z i u m = \frac{1}{2} \times$
$(s u m o f p a r a l l e l s i d e s) \times h e i g h t]$
$= \frac{1}{2} (1 + 2.5 + 1.5) = \frac{1}{2} \times 5 = \frac{5}{2} m$

PHXI03:MOTION IN A STRAIGHT LINE

362316 A body travels $102.5 m$ in $n^{th}$ second and $115.0 m$ in $(n + 2)^{th}$ second. The acceleration is

1

9 m / s^{2}

2

6.25 m / s^{2}

3

5 m / s^{2}

4

12.5 m / s^{2}

Explanation:

Distance travelled in $n^{th}$ second is
Use $S_{n^{t h}} = u + \frac{1}{2} a (2 n - 1)$
$102.5 = u + \frac{1}{2} a (2 n - 1)$ $(1)$
$115 = u + \frac{1}{2} a (2 \times (n + 2) - 1)$
$115 = u + \frac{1}{2} a (2 n + 3) (2)$
Subtract equation (1) from (2), we get
$115 - 102.5 = \frac{1}{2} a (2 n + 3 - 2 n + 1)$
$\Rightarrow a = 6.25 m / s^{2}$

JEE - 2024

PHXI03:MOTION IN A STRAIGHT LINE

362317 An electron starting from rest has a velocity that increases linearly with time as $v = k t$ , where $k = 2 m / s^{2}$ . The distance covered in the first
3 seconds will be :

1 9

m

2 16

m

3 27

m

4 36

m

Explanation:

Variation of velocity with $t$ is given as $v = k t$
$\Rightarrow \frac{d x}{d t} = k t$
$\Rightarrow d x = k t d t$
Distance covered in first 3 seconds will be
$\int_{0}^{x} d x = k \int_{0}^{3} t d t$
$\Rightarrow x = \frac{1}{2} k {[t^{2}]}_{0}^{3} = \frac{1}{2} \times 2 \times 9$
$∴ x = 9 m$ .
So correct option is (1)

PHXI03:MOTION IN A STRAIGHT LINE

362318 A car starting from rest accelerates at the rate $f$ through a distance $s$ , then continues at a constant speed for time $t$ and then decelerates at the rate $f / 2$ to come to rest. If the total time taken is $15 s$ , then

1

s = f t

2

s = \frac{1}{6} f t^{2}

3

s = \frac{1}{72} f t^{2}

4

s = \frac{1}{4} f t^{2}

Explanation:

supporting img
$v = \sqrt{2 f s} \Rightarrow 12 s = \sqrt{2 f s t}$
$\Rightarrow 144 s^{2} = 2 f s t^{2}$
$\Rightarrow s = \frac{1}{72} f t^{2}$

AIIMS - 2008

PHXI03:MOTION IN A STRAIGHT LINE

362315 The speed-time graph of a particle moving along a solid curve is shown below. The distance traversed by the particle from $t = 0 s$ to $t = 3 s$ is
supporting img

1

5 m

2

\frac{5}{2} m

3

\frac{10}{3} m

4

2 m

Explanation:

supporting img
Distance travelled by the particle ,
$s =$ Area under the curve
$= s_{01} + s_{12} + s_{23}$
$s =$ Area of $△ O A B$
+ Area of trapezium A B C D
$+ A r e a o f Δ D C E$
$= \frac{1}{2} \times O B \times A B + \frac{1}{2} (A B + C D) \times B D + \frac{1}{2}$
$\times D E \times D C$
$= \frac{1}{2} \times (1 - 0) \times 1 + \frac{1}{2} \times (1 + 1.5) \times (2 - 1)$
$+ \frac{1}{2} \times (3 - 2) \times 1.5$
$= \frac{1}{2} + \frac{1}{2} \times 2.5 + \frac{1}{2} \times 1.5$
[ $∵$ area of triangle $= \frac{1}{2} \times$ base $\times$ height]
$a r e a o f t r a p e z i u m = \frac{1}{2} \times$
$(s u m o f p a r a l l e l s i d e s) \times h e i g h t]$
$= \frac{1}{2} (1 + 2.5 + 1.5) = \frac{1}{2} \times 5 = \frac{5}{2} m$

PHXI03:MOTION IN A STRAIGHT LINE

362316 A body travels $102.5 m$ in $n^{th}$ second and $115.0 m$ in $(n + 2)^{th}$ second. The acceleration is

1

9 m / s^{2}

2

6.25 m / s^{2}

3

5 m / s^{2}

4

12.5 m / s^{2}

Explanation:

Distance travelled in $n^{th}$ second is
Use $S_{n^{t h}} = u + \frac{1}{2} a (2 n - 1)$
$102.5 = u + \frac{1}{2} a (2 n - 1)$ $(1)$
$115 = u + \frac{1}{2} a (2 \times (n + 2) - 1)$
$115 = u + \frac{1}{2} a (2 n + 3) (2)$
Subtract equation (1) from (2), we get
$115 - 102.5 = \frac{1}{2} a (2 n + 3 - 2 n + 1)$
$\Rightarrow a = 6.25 m / s^{2}$

JEE - 2024

PHXI03:MOTION IN A STRAIGHT LINE

362317 An electron starting from rest has a velocity that increases linearly with time as $v = k t$ , where $k = 2 m / s^{2}$ . The distance covered in the first
3 seconds will be :

1 9

m

2 16

m

3 27

m

4 36

m

Explanation:

Variation of velocity with $t$ is given as $v = k t$
$\Rightarrow \frac{d x}{d t} = k t$
$\Rightarrow d x = k t d t$
Distance covered in first 3 seconds will be
$\int_{0}^{x} d x = k \int_{0}^{3} t d t$
$\Rightarrow x = \frac{1}{2} k {[t^{2}]}_{0}^{3} = \frac{1}{2} \times 2 \times 9$
$∴ x = 9 m$ .
So correct option is (1)

PHXI03:MOTION IN A STRAIGHT LINE

362318 A car starting from rest accelerates at the rate $f$ through a distance $s$ , then continues at a constant speed for time $t$ and then decelerates at the rate $f / 2$ to come to rest. If the total time taken is $15 s$ , then

1

s = f t

2

s = \frac{1}{6} f t^{2}

3

s = \frac{1}{72} f t^{2}

4

s = \frac{1}{4} f t^{2}

Explanation:

supporting img
$v = \sqrt{2 f s} \Rightarrow 12 s = \sqrt{2 f s t}$
$\Rightarrow 144 s^{2} = 2 f s t^{2}$
$\Rightarrow s = \frac{1}{72} f t^{2}$

AIIMS - 2008