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MOTION IN A STRIGHT LINE

269772 A car moving with a speed of $50 km / hr$ can be stopped by brakes after atleast $6 m$ . If the same car is moving at a speed of $100 km / hr$ , theminimum stopping distance is

1

12 m

2

18 m

3

24 m

4

6 m

Explanation:

$v^{2} - u^{2} = 2 a s, v = 0; u^{2} \propto s$

MOTION IN A STRIGHT LINE

269773 A particle moving with a constant acceleration describes in the last second of its motion $36 %$ of the whole distance. If it starts from rest, how long is the particle in motion and through what distance does it moves if it describes $6 cm$ in the first sec.?

1

5 s; 150 cm

2

10 s; 150 cm

3

15 s; 100 cm

4

20 s; 200 cm

Explanation:

$s = \frac{1}{2} a t^{2}; \frac{2 n - 1}{n^{2}} = \frac{36}{100}$

MOTION IN A STRIGHT LINE

269774 Abusstarts from rest with a constant acceleration of $5 m / s^{2}$ .At the same time a car travelling with a constant velocity $50 m / sover$ takes and passes the bus. How fast is the bus travelling when they are side by side?

1

10 m / s

2

50 m / s

3

100 m / s

4

150 m / s

Explanation:

$s = \frac{1}{2} a t^{2}, s = v t, v = d t$

MOTION IN A STRIGHT LINE

269775 A particle moving with uniform retardation covers distances $18 m, 14 m$ and $10 m$ in successive seconds. It comes to rest after travelling a further distance of

1

50 m

2

8 m

3

12 m

4

42 m

Explanation:

$s = u + a (n - \frac{1}{2})$

MOTION IN A STRIGHT LINE

269796 The relation betweentimet and distance $x$ is $t = a x^{2} + b x$ wherea and $b$ areconstants. The acceleration is

1

- 2 a v^{3}

2

2 a v^{2}

3

- 2 a b v^{2}

4

2 b v^{3}

Explanation:

$v = \frac{d x}{d t}, a = \frac{d v}{d t}$

MOTION IN A STRIGHT LINE

269772 A car moving with a speed of $50 km / hr$ can be stopped by brakes after atleast $6 m$ . If the same car is moving at a speed of $100 km / hr$ , theminimum stopping distance is

1

12 m

2

18 m

3

24 m

4

6 m

Explanation:

$v^{2} - u^{2} = 2 a s, v = 0; u^{2} \propto s$

MOTION IN A STRIGHT LINE

269773 A particle moving with a constant acceleration describes in the last second of its motion $36 %$ of the whole distance. If it starts from rest, how long is the particle in motion and through what distance does it moves if it describes $6 cm$ in the first sec.?

1

5 s; 150 cm

2

10 s; 150 cm

3

15 s; 100 cm

4

20 s; 200 cm

Explanation:

$s = \frac{1}{2} a t^{2}; \frac{2 n - 1}{n^{2}} = \frac{36}{100}$

MOTION IN A STRIGHT LINE

269774 Abusstarts from rest with a constant acceleration of $5 m / s^{2}$ .At the same time a car travelling with a constant velocity $50 m / sover$ takes and passes the bus. How fast is the bus travelling when they are side by side?

1

10 m / s

2

50 m / s

3

100 m / s

4

150 m / s

Explanation:

$s = \frac{1}{2} a t^{2}, s = v t, v = d t$

MOTION IN A STRIGHT LINE

269775 A particle moving with uniform retardation covers distances $18 m, 14 m$ and $10 m$ in successive seconds. It comes to rest after travelling a further distance of

1

50 m

2

8 m

3

12 m

4

42 m

Explanation:

$s = u + a (n - \frac{1}{2})$

MOTION IN A STRIGHT LINE

269796 The relation betweentimet and distance $x$ is $t = a x^{2} + b x$ wherea and $b$ areconstants. The acceleration is

1

- 2 a v^{3}

2

2 a v^{2}

3

- 2 a b v^{2}

4

2 b v^{3}

Explanation:

$v = \frac{d x}{d t}, a = \frac{d v}{d t}$

MOTION IN A STRIGHT LINE

269772 A car moving with a speed of $50 km / hr$ can be stopped by brakes after atleast $6 m$ . If the same car is moving at a speed of $100 km / hr$ , theminimum stopping distance is

1

12 m

2

18 m

3

24 m

4

6 m

Explanation:

$v^{2} - u^{2} = 2 a s, v = 0; u^{2} \propto s$

MOTION IN A STRIGHT LINE

269773 A particle moving with a constant acceleration describes in the last second of its motion $36 %$ of the whole distance. If it starts from rest, how long is the particle in motion and through what distance does it moves if it describes $6 cm$ in the first sec.?

1

5 s; 150 cm

2

10 s; 150 cm

3

15 s; 100 cm

4

20 s; 200 cm

Explanation:

$s = \frac{1}{2} a t^{2}; \frac{2 n - 1}{n^{2}} = \frac{36}{100}$

MOTION IN A STRIGHT LINE

269774 Abusstarts from rest with a constant acceleration of $5 m / s^{2}$ .At the same time a car travelling with a constant velocity $50 m / sover$ takes and passes the bus. How fast is the bus travelling when they are side by side?

1

10 m / s

2

50 m / s

3

100 m / s

4

150 m / s

Explanation:

$s = \frac{1}{2} a t^{2}, s = v t, v = d t$

MOTION IN A STRIGHT LINE

269775 A particle moving with uniform retardation covers distances $18 m, 14 m$ and $10 m$ in successive seconds. It comes to rest after travelling a further distance of

1

50 m

2

8 m

3

12 m

4

42 m

Explanation:

$s = u + a (n - \frac{1}{2})$

MOTION IN A STRIGHT LINE

269796 The relation betweentimet and distance $x$ is $t = a x^{2} + b x$ wherea and $b$ areconstants. The acceleration is

1

- 2 a v^{3}

2

2 a v^{2}

3

- 2 a b v^{2}

4

2 b v^{3}

Explanation:

$v = \frac{d x}{d t}, a = \frac{d v}{d t}$

MOTION IN A STRIGHT LINE

269772 A car moving with a speed of $50 km / hr$ can be stopped by brakes after atleast $6 m$ . If the same car is moving at a speed of $100 km / hr$ , theminimum stopping distance is

1

12 m

2

18 m

3

24 m

4

6 m

Explanation:

$v^{2} - u^{2} = 2 a s, v = 0; u^{2} \propto s$

MOTION IN A STRIGHT LINE

269773 A particle moving with a constant acceleration describes in the last second of its motion $36 %$ of the whole distance. If it starts from rest, how long is the particle in motion and through what distance does it moves if it describes $6 cm$ in the first sec.?

1

5 s; 150 cm

2

10 s; 150 cm

3

15 s; 100 cm

4

20 s; 200 cm

Explanation:

$s = \frac{1}{2} a t^{2}; \frac{2 n - 1}{n^{2}} = \frac{36}{100}$

MOTION IN A STRIGHT LINE

269774 Abusstarts from rest with a constant acceleration of $5 m / s^{2}$ .At the same time a car travelling with a constant velocity $50 m / sover$ takes and passes the bus. How fast is the bus travelling when they are side by side?

1

10 m / s

2

50 m / s

3

100 m / s

4

150 m / s

Explanation:

$s = \frac{1}{2} a t^{2}, s = v t, v = d t$

MOTION IN A STRIGHT LINE

269775 A particle moving with uniform retardation covers distances $18 m, 14 m$ and $10 m$ in successive seconds. It comes to rest after travelling a further distance of

1

50 m

2

8 m

3

12 m

4

42 m

Explanation:

$s = u + a (n - \frac{1}{2})$

MOTION IN A STRIGHT LINE

269796 The relation betweentimet and distance $x$ is $t = a x^{2} + b x$ wherea and $b$ areconstants. The acceleration is

1

- 2 a v^{3}

2

2 a v^{2}

3

- 2 a b v^{2}

4

2 b v^{3}

Explanation:

$v = \frac{d x}{d t}, a = \frac{d v}{d t}$

MOTION IN A STRIGHT LINE

269772 A car moving with a speed of $50 km / hr$ can be stopped by brakes after atleast $6 m$ . If the same car is moving at a speed of $100 km / hr$ , theminimum stopping distance is

1

12 m

2

18 m

3

24 m

4

6 m

Explanation:

$v^{2} - u^{2} = 2 a s, v = 0; u^{2} \propto s$

MOTION IN A STRIGHT LINE

269773 A particle moving with a constant acceleration describes in the last second of its motion $36 %$ of the whole distance. If it starts from rest, how long is the particle in motion and through what distance does it moves if it describes $6 cm$ in the first sec.?

1

5 s; 150 cm

2

10 s; 150 cm

3

15 s; 100 cm

4

20 s; 200 cm

Explanation:

$s = \frac{1}{2} a t^{2}; \frac{2 n - 1}{n^{2}} = \frac{36}{100}$

MOTION IN A STRIGHT LINE

269774 Abusstarts from rest with a constant acceleration of $5 m / s^{2}$ .At the same time a car travelling with a constant velocity $50 m / sover$ takes and passes the bus. How fast is the bus travelling when they are side by side?

1

10 m / s

2

50 m / s

3

100 m / s

4

150 m / s

Explanation:

$s = \frac{1}{2} a t^{2}, s = v t, v = d t$

MOTION IN A STRIGHT LINE

269775 A particle moving with uniform retardation covers distances $18 m, 14 m$ and $10 m$ in successive seconds. It comes to rest after travelling a further distance of

1

50 m

2

8 m

3

12 m

4

42 m

Explanation:

$s = u + a (n - \frac{1}{2})$

MOTION IN A STRIGHT LINE

269796 The relation betweentimet and distance $x$ is $t = a x^{2} + b x$ wherea and $b$ areconstants. The acceleration is

1

- 2 a v^{3}

2

2 a v^{2}

3

- 2 a b v^{2}

4

2 b v^{3}

Explanation:

$v = \frac{d x}{d t}, a = \frac{d v}{d t}$