Question

the velocity-time graph shows the motion of a cyclist. find (a) its acceleration (b)velocity (c)distance covered by the cyclist in 15 seconds.​

Answer 1

Answer:

where is the velocity time graph

Answer 2

Its acceleration - 0 ms²

Velocity - 20 ms⁻¹

The distance covered by the cyclist in 15 seconds - 300 m.

Accurate Question :          [ With Graph ]

The velocity - time graph shows the motion of a cyclist. Find :

(i) Its Acceleration.

(ii) Its Velocity after 20 seconds.

(iii) The distance covered by the cyclist in 15 seconds.

                [ Refer the Attachment for the graph ]

Explanation :

We have only given a velocity time graph.

So,

To Find :

• Acceleration .
• Velocity
• Distance covered in 15 seconds.

$\rule{400}{1.5}$

• Acceleration :

We know that the slope of velocity time graph gives the acceleration of a given body.

So,

$\bigstar\;\boxed{ \sf{The\;Slope = \dfrac{v2 - v1}{t2 - t1} = \bf \tan( \theta) = a}}}$

Let's consider the :

$\bf v_2 = Final\;velocity\\ \\ \bf v_1= Initial\;velocity$

Hence,

$\sf Time\;interval = t_2-t_1$

We are given,

$\implies \sf v_1 = 20 \: ms {}^{ - 1} }}\\ \\ \\\implies \sf v_2= 20 \: m {s}^{- 1} }} \\ \\ \\ \\implies \sf {t_2 - t_1 = 25 \; sec}}$

We know that :

$\bigstar\;\boxed{\bf a = \dfrac{v_0-v}{T}}}$

So,

$\implies \sf a = \dfrac{20 - 20}{25} = 0 \: m {s}^{-2}}}$

Therefore, The acceleration will be 0 ms²

$\rule{400}{1.5}$

• Velocity :

Here,

The magnitude remains constant, so the body will move with constant velocity. The graph is also a straight line parallel to time axis.

Therefore, The velocity will be 20 ms⁻¹

$\rule{400}{1.5}$

• Distance in first 15 seconds :

When the given velocity is constant (uniform) :

So,

$\bigstar\;\boxed{\bf Distance= ab \times ad}}}$

Hence,

Distance :

⇒ 20 × 15  m

⇒ s = 300 m.

Hence,

Its acceleration - 0 ms²

Velocity - 20 ms⁻¹

The distance covered by the cyclist in 15 seconds - 300 m.