Question

026. The velocity time graph in the figure shows the motion of
cyclist
Find the following
It's acceleration
b. it's velocity
c. The distance covered by the cyclist in 15 seconds.​

Answer 1

Answer:

Given:

Velocity vs Time graph has been provided as on the photo. It shows a linear graph parallel to the x axis.

To find:

• Acceleration
• Velocity
• Distance travelled in 15 seconds

Calculation:

For Acceleration:

The velocity vs time graph is linear and parallel to time axis . This means slope of this graph is zero as there is no change in velocity wrt time. Hence

Acceleration is zero.

For velocity:

As per the graph, the object at a constant velocity of 20 m/s

Distance travelled:

Distance travelled is given by area under velocity vs time graph

Distance be d :

$d = area \: of \: graph$

$= > d = 20 \times 15$

$= > d = 300 \: m {s}^{ - 1}$

Answer 2

✯Answer✯

❋Given:

• A velocity time graph

❋To find:

• Accleration
• Velocity
• Distance covered in 15 seconds.

$\large{ \bold{ \red{ \underline{accleration}}}}$

The slope of velocity time graph gives the accleration of a given body.

$\large{ \boxed{ \sf{slope = \frac{v2 - v1}{t2 - t1} = \tan( \theta) = a}}}$

☛Let for the above case v2 = final velocity

v1= initial velocity

t2-t1 = time interval.

☛ Here,

$\large{ \sf{v1 = 20 \: ms {}^{ - 1} }} \\ \\ \large{ \sf{v2 = 20 \: m {s}^{ - 1} }} \\ \\ \large{ \sf{t2 - t1 = 25 - 0 = 25 \: sec}}$

☛By using formula

$\large{ \sf{ \boxed{ \green{a = \frac{20 - 20}{25} = 0 \: m {s}^{ - 2} }} }}$

$\large{ \sf{ \underline{ \red{velocity}}}}$

Here, the body moves with constant velocity, as the magnitude remains constant.

☛The graph is a stratight line parallel to time axis.

$\large{ \sf{ \boxed{ \green{v = 20 \: m {s}^{ - 1} }}}}$

$\large{ \sf{ \underline{ \red{distance \: in \: first \: 15 \: seconds}}}}$

As we can say, Velocity × time = Displacement ( here distance) only, We can calculate the distance from the above graph, by finding area under graph.

☛When velocity is constant(uniform)

$\large{ \boxed{ \sf{distance = area \: of \: triangle \: abcd = ab \times ad}}}$

▶️Refer to the attachment

☛By using formula,

$\large{ \sf{ distance = 20 \times 15 \: m}} \\ \\ \large{ \boxed{ \green{s = 300m}}}$

✯So final Answer ✯

$\large{ \boxed{ \bf{a = 0 \: m {s}^{ - 2} }}} \\ \\ \large{ \boxed{ \bf{v = 20 \: m {s}^{ - 1} }}} \\ \\ \large{ \boxed{ \bf{s = 300m}}}$