Question

pls help me in solving this graph​

Answer 1

Answer:

(a) 80 m/s and 8b time

(b) 60 m/s

(c) d and e

(d) car A and B

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Answer 2

19)

a)

Given:

According to graph,

For the time interval of 0 to 8 s

Initial velocity of car A, u= 0 m/s

Final velocity of car A, v= 80 m/s

Time taken by car, t= 8 s

To Find:

Acceleration of car A between 0 to 8 s

Solution:

We know that,

• For a body moving in uniform motion or having constant acceleration, acceleration 'a' of body is given by

$\pink{\boxed{\bf{a=\dfrac{v-u}{t}}}}$

where,

a is acceleration

v is final velocity

u is initial velocity

t is time taken

$\rule{190}{1}$

Let the acceleration of car A between 0 to 8 s be 'a'

So,

$\longrightarrow\rm{a=\dfrac{v-u}{t}}$

$\longrightarrow\rm{a=\dfrac{80-0}{8}}$

$\longrightarrow\rm{a=\dfrac{\cancel{80}}{\cancel{8}}}$

$\longrightarrow\rm\green{a=10\:m/s^{2}}$

Hence, the acceleration of car A between 0 to 8 s is 10 m/s².

$\rule{190}{2}$

b)

Given:

According to graph,

For the time interval of 2s to 4s

Initial velocity of car B, u= 20 m/s

Final velocity of car B, v= 60 m/s

Time taken by car B, t= 2 s

To Find:

Acceleration of car B between 2 to 4 s

Solution:

Let the acceleration of car B between 2 to 4 s be 'a'

So,

$\longrightarrow\rm{a=\dfrac{v-u}{t}}$

$\longrightarrow\rm{a=\dfrac{60-20}{2}}$

$\longrightarrow\rm{a=\dfrac{\cancel{40}}{\cancel{2}}}$

$\longrightarrow\rm\green{a=20\:m/s^{2}}$

Hence, the acceleration of car B between 2 to 4 s is 20 m/s².

$\rule{190}{2}$

c)

The points of time at which both cars have same velocity can be determined by the help of graph in the following way-

• Firstly, we have to determine the points at which y- coordinates of both line is same or points at which both the lines are intersecting
• Secondly, we have to find the time corresponding to the intersection points

So,

From the graph, it is clear that ,these lines are intersecting where the velocity of both the cars is 20 m/s and 60m/s

Therefore, time corresponding to above velocities are 2 s and 6 s.

Hence, at 2 s and 6 s both cars have same velocity.

$\rule{190}{2}$

d)

To Find:

Which car will be ahead after 8 s and by how much

Solution:

• According to second equation of motion for constant acceleration,

$\orange{\boxed{\bf{s=ut+\dfrac{1}{2}at^{2}}}}$

where,

u is initial velocity

a is acceleration

s is displacement

t is time taken

$\rule{190}{1}$

For Car A

Between 0 to 8 s

Initial velocity, u= 0 m/s

Acceleration, a= 10 m/s²

Time taken, t= 8 s

Let s be the displacement of car A in first 8 s

On applying second equation of motion on car A, we get

$\longrightarrow\rm{s=ut+\dfrac{1}{2}at^{2}}$

$\longrightarrow\rm{s=0(8)+\dfrac{1}{2}(10)(8)^{2}}$

$\longrightarrow\rm{s=\dfrac{1}{2}(10)(64)}$

$\longrightarrow\rm{s=\dfrac{640}{2}}$

$\longrightarrow\rm\green{s=320\:m}$

$\rule{190}{1}$

For Car B

Since, motion of car B is not uniform in first 8 s, so we have to consider the case

Case-1: Between 0 to 1 s

Initial velocity, u₁= 0 m/s

Final velocity, v₁= 0 m/s

Time taken, t₁= 1 s

Let the acceleration of car B for given interval be a₁

So,

$\longrightarrow\rm{a_{1}=\dfrac{0-0}{1}=0\:m/s^{2}}$

Let s₁ be the displacement of car B from 0 to 1 s

On applying second equation of motion on car B, we get

$\longrightarrow\rm{s_{1}=u_{1}t_{1}+\dfrac{1}{2}a_{1}t_{1}^{2}}$

$\longrightarrow\rm{s_{1}=0(1)+\dfrac{1}{2}(0)(1)^{2}}$

$\longrightarrow\rm{s_{1}=0\:m}$

Case-2: Between 1 to 4 s

Initial velocity, u₂= 0 m/s

Final velocity, v₂= 60 m/s

Time taken, t₂= 3 s

Let the acceleration of car B for given interval be a₂

So,

$\longrightarrow\rm{a_{2}=\dfrac{60-0}{3}}$

$\longrightarrow\rm{a_{2}=\dfrac{60}{3}=20\:m/s^{2}}$

Let s₂ be the displacement of car B from 1 to 4 s

On applying second equation of motion on car B, we get

$\longrightarrow\rm{s_{2}=u_{2}t_{2}+\dfrac{1}{2}a_{2}t_{2}^{2}}$

$\longrightarrow\rm{s_{2}=0(3)+\dfrac{1}{2}(20)(3)^{2}}$

$\longrightarrow\rm{s_{2}=\dfrac{1}{2}(20)(9)}$

$\longrightarrow\rm{s_{2}=\dfrac{180}{2}}$

$\longrightarrow\rm{s_{2}=90\:m}$

Case-3: Between 4 to 8 s

Initial velocity, u₃= 60 m/s

Final velocity, v₃= 60 m/s

Time taken, t₃= 4 s

Let the acceleration of car B for given interval be a₃

So,

$\longrightarrow\rm{a_{3}=\dfrac{60-60}{4}=0\:m/s^{2}}$

Let s₃ be the displacement of car B from 4 to 8 s

On applying second equation of motion on car B, we get

$\longrightarrow\rm{s_{3}=u_{3}t_{3}+\dfrac{1}{2}a_{3}t_{3}^{2}}$

$\longrightarrow\rm{s_{3}=60(4)+\dfrac{1}{2}(0)(4)^{2}}$

$\longrightarrow\rm{s_{3}=60(4)}$

$\longrightarrow\rm{s_{3}=240\:m}$

Now, let the total distance covered be car B be s'

So,

$\longrightarrow\rm{s'=s_{1}+s_{2}+s_{3}}$

$\longrightarrow\rm{s'=0+90+240}$

$\longrightarrow\rm\green{s'=330\:m}$

$\rule{190}{1}$

Now, we will calculate s'-s

$\longrightarrow\rm\green{s'-s=330-320=10\:m}$

Hence, after 8 s Car B will be ahead of Car A by 10 m.