Question

A bike is initially moving at 14 m/s in a straight line. It then accelerates uniformly in
the same direction for one fourth of a second to a velocity of 18 m/s. How much
distance does the bike cover?

Answer 1

Question :-

A bike is initially moving at 14 m/s in a straight line. It then accelerates uniformly in the same direction for one fourth of a second to a velocity of 18 m/s. How much distance does the bike cover?

Given :-

• Initial Velocity, u = 14m/s
• Final Velocity, v = 18m/s
• Time, t = $\dfrac {1}{4}$s

To Find :-

• Distance covered by bike.

Solution :-

To find the distance, we have to firstly find acceleration.

So,

Acceleration, a = $\dfrac {v \: - \: y}{t}$

$\implies$ $\dfrac {18 \: - \: 14}{1/4}$

$\implies$ $\dfrac {4 \: × \: 4}{1}$

$\implies$ $\dfrac {16}{1}$

$\implies$ $16m/s^2$

Now,

According to the second law of motion,

$\boxed {s \: = \: ut \: + \: 1/2at^2}$

Therefore,

$s \: = \: (12×1/4) \: + \: 1/2×16×{1/4}^2$

$s \: = \: 12/4 \: + \: 1/2×16×1/16$

$s \: = \: 12/4 \: + \: 1/2$

$s \: = \: 3.5m$

Answer 2

Answer:

The distance covered by the bike is 4 m.

Explanation:

Given data :

initial velocity (u) = 14 m/s

Time (t) = 1/4 seconds

final velocity (v) = 18 m/s

Data to be calculated :

Distance covered by the bike

Formula to be used :

$a = \frac{v - u}{t}$

where a is the acceleration of the bike

$s \: = ut \: + \frac{1}{2} a {t}^{2}$

Obtaining Results :

$a \: = \frac{18 - 14}{ \frac{1}{4} } = 4(4) = 16 \: m {s }^{ - 2}$

Now using Newton's 2nd equation of motion :

$s \: = 14 \times \frac{1}{4} + \frac{1}{2} \times 16 \times { \frac{1}{4} }^{2}$

$s \: = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4m$

Hence, The distance covered by the bike is 4 m.

Concept :

There are three equations of motion :

$v \: = u \: + at \\ s = ut + \frac{1}{2} a {t}^{2} \\ {v}^{2} - {u}^{2} = 2as$

These equations can be used only when the acceleration of the body moving is constant.

Using these we can solve our questions easily.

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