Question

The velocity−time graph for the motion of a car is shown in the given figure.
(a) What is the total distance travelled by the car?
(b) What is the speed of the car (in km/h) at 8th second?
(c) What is the acceleration of the car
(i) during the first 6 s?
(ii) during the last 2 s?
I will mark u brainliest . Please help

Answer 1

Solving

(a) What is the total distance travelled by the car?

• take, Distance = s

Formula

$\:$

$\: \: \large \rm \mapsto \boxed{ \rm\: \: s = ut + \dfrac{1}{2} a {t}^{2} }$

$\:$

Here,

• u = Initial Velocity

• t = time

• a = Acceleration

$\:$

$\: \: \large \rm \mapsto \boxed{ \rm\: \: a = \frac{v - u}{t} }$

$\:$

here,

• v = Final Velocity = 15m

• u = 5m

• t = 10s

$\:$

$\large \rm \: \: \: \: : \implies{ \rm\: \: a = \dfrac{15 - 5}{10} }$

$\:$

$\large \rm \: \: \: \: : \implies{ \rm\: \: a = \dfrac{10}{10} }$

$\:$

$\large \rm \: \: \: \: : \implies{ \rm\: \: a = \dfrac{ \cancel{10}}{ \cancel{10} }}$

$\:$

$\large \rm \therefore{ \rm\: \: a = 1m/s}$

$\:$

$\: \: \large \rm \mapsto \boxed{ \rm\: \: s = ut + \dfrac{1}{2} a {t}^{2} }$

$\:$

$\large \rm \: \: \: : \implies{ \rm\: \: s = 5 \times 10 + \dfrac{1}{2}(1)( {10)}^{2} }$

$\:$

$\large \rm \: \: \: : \implies{ \rm\: \: s = 50 + \dfrac{1}{2} \times 10 \times 10 }$

$\:$

$\large \rm \: \: \: : \implies{ \rm\: \: s = 50 + \dfrac{1}{ \cancel2} \times \cancel{10} \: \:5 \times 10 }$

$\:$

$\large \rm \: \: \: : \implies{ \rm\: \: s = 50 +50 }$

$\:$

∴ Distance = 100m

$\:$

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(b) What is the speed of the car (in km/h) at 8th second?

$\:$

Formula

$\:$

$\large \: \: \: \: \mapsto \: \boxed{ \rm \: speed = \frac{distance}{time} }$

$\:$

For that,

$\: \: \large \rm \mapsto \boxed{ \rm\: \: s = ut + \dfrac{1}{2} a {t}^{2} }$

$\:$

Here,

• u = 5m

• t = 8s

• a = ?

$\: \: \large \rm \mapsto \boxed{ \rm\: \: a = \frac{v - u}{t} }$

$\:$

$\large \rm : \implies{ \rm\: \: a = \dfrac{15 - 5}{8} }$

$\:$

$\large \rm \therefore{ \rm\: \: a = \dfrac{10}{8} }$

$\:$

• Finding distance

$\:$

$\large \rm : \implies{ \rm\: \: s = 5 \times 8+ \dfrac{1}{2} \left( \dfrac{10}{8} \right)( {10)}^{2} }$

$\:$

$\large \rm : \implies{ \rm\: \: s = 40+ \dfrac{1}{ \cancel2} \left( \dfrac{ \cancel{10} \: \: 5}{8} \right)( {10)}^{2} }$

$\:$

$\large \rm : \implies{ \rm\: \: s = 40+ \dfrac{5}{8} \times 10 \times 10 }$

$\:$

$\large \rm : \implies{ \rm\: \: s = 40+ \dfrac{5}{ \cancel8 \: \: 4} \times \cancel{10} \: \: 5 \times 10 }$

$\:$

$\large \rm : \implies{ \rm\: \: s = 40+ \dfrac{5}{ \cancel4 \: \: 2} \times 5 \times \cancel{10 } \: \: 5}$

$\:$

$\large \rm : \implies{ \rm\: \: s = 40+ \dfrac{5}{ 2} \times 5 \times 5}$

$\:$

$\large \rm : \implies{ \rm\: \: s = 40+ \dfrac{5}{ 2} \times 25}$

$\:$

$\large \rm : \implies{ \rm\: \: s = 40+ \dfrac{125}{ 2} }$

$\:$

$\large \rm : \implies{ \rm\: \: s = \dfrac{80 + 125}{ 2} }$

$\:$

$\large \rm : \implies{ \rm\: \: s = \dfrac{205}{ 2} }$

$\:$

$\large \: \: \: \: \mapsto \: \boxed{ \rm \: speed = \frac{distance}{time} }$

$\:$

$\large \rm : \implies{ \rm\: \: speed = \dfrac{205}{ 2 \times 8} }$

$\:$

$\large \rm : \implies{ \rm\: \: speed = \dfrac{205}{ 16} }$

$\:$

$\large \rm \therefore{ \rm\: \: speed = 12.8m /s}$

$\:$

Speed = 0.0128km/h

$\:$

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$\:$

Answer 2

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To Find

(a) What is the total distance travelled by the car?

(b) What is the speed of the car (in km/h) at 8th second?

(c) What is the acceleration of the car

(i) During the first 6 s

(ii) During the last 2 s

Solution

Calculating acceleration of car from 0-6 s,

➝ v = u + at

➝ 15 = 5 + 6a

➝ 15 - 5 = 6a

➝ 6a = 10

➝ a = 10/6

After cancelling, we get :-

➝ a = 1.67 m/s²

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◈ (a)Calculating distance travelled in 0-6 s,

➳ s = ut + 1/2 at²

➳ s = 5 × 6 + 1/2 × 1.67 × 36

➳ s = 30 + 30

➳ s = 60 m

Calculating distance travelled in 6-10 s,

➪ Distance = Speed × Time

➪ Distance = 15 × 4

➪ Distance = 60 m

So, total distance :-

⪼ 60 + 60

⪼ 120 m

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◈ (b)At 8th s, car us moving with velocity 15 m/s

So,

→ Speed in km/h = 15 × 18/5\:

→ Speed in km/h = 54 km/h

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◈ (c)Acceleration of car during 1st 6s is calculated above, i.e, 1.67 m/s²

During last 2 s, car us moving with constant velocity of 15 m/s. So, it's acceleration is 0 m/s²

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