Question

(а) Пеге 13 пU want to come to rest. 1. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is (Hint: 1 tonne = 1000 kg.)​

Answer 1

Answer:

Given data:

Given data:Mass of stone(m)=7000 kg

Given data:Mass of stone(m)=7000 kgDistance(S)=400 meter

Given data:Mass of stone(m)=7000 kgDistance(S)=400 meterTime(t)=20 second

Given data:Mass of stone(m)=7000 kgDistance(S)=400 meterTime(t)=20 second1) We have to find the acceleration of the truck .

$\frac{1}{2 \: } a \times {t }^{2} \: where \: u \: = 0 \\ s = \frac{1}{2}a \times t {}^{2} \\ \\ a = \frac{2 \times s}{ {t}^{2} } \: = \frac{2 \times 400}{2 0\times 20} \: \\ = 2 \times \frac{m}{ {sec} ^{2} } \\ \\ \\ Force \ \: acting \: on \: truck \: \\ </em></p><p></p><p><em>[tex] \frac{1}{2 \: } a \times {t }^{2} \: where \: u \: = 0 \\ s = \frac{1}{2}a \times t {}^{2} \\ \\ a = \frac{2 \times s}{ {t}^{2} } \: = \frac{2 \times 400}{2 0\times 20} \: \\ = 2 \times \frac{m}{ {sec} ^{2} } \\ \\ \\ Force \ \: acting \: on \: truck \: \\ So,from \: newton's  \: \\ 2nd \:  law \: of \: motion \\ </em></p><p><em>[tex] \frac{1}{2 \: } a \times {t }^{2} \: where \: u \: = 0 \\ s = \frac{1}{2}a \times t {}^{2} \\ \\ a = \frac{2 \times s}{ {t}^{2} } \: = \frac{2 \times 400}{2 0\times 20} \: \\ = 2 \times \frac{m}{ {sec} ^{2} } \\ \\ \\ Force \ \: acting \: on \: truck \: \\ So,from \: newton's  \: \\ 2nd \:  law \: of \: motion \\ F = ma = 7000 \times 2 \\ = 14000n = 14kn</em></p><p><em>[tex] \frac{1}{2 \: } a \times {t }^{2} \: where \: u \: = 0 \\ s = \frac{1}{2}a \times t {}^{2} \\ \\ a = \frac{2 \times s}{ {t}^{2} } \: = \frac{2 \times 400}{2 0\times 20} \: \\ = 2 \times \frac{m}{ {sec} ^{2} } \\ \\ \\ Force \ \: acting \: on \: truck \: \\ So,from \: newton's  \: \\ 2nd \:  law \: of \: motion \\ F = ma = 7000 \times 2 \\ = 14000n = 14kn$

Answer 2

To Find :

• The acceleration of the truck
• Force acting on the truck

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Given :

• Initial velocity , u = 0m/s
• Distance travelled , s = 400 m
• Time taken,t = 20 s

Note : You have not provided mass, so I will take it as 10 tonnes . Please remember to provide everything from next time

• mass = 10 tonnes

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Formula to be used :

• Distance = ½(u+v)t
• Acceleration = (v-u)/t
• Force = mass×acceleration

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Solution :

We are given , the initial velocity, distance and time .

Hence, we need to find Final velocity and acceleration .

We know that,

s = ½(v+u)t

$\implies \tt{400 = \frac{1}{2} \times (0 + v)20}$

$\implies \tt{20v = 400 \times 2}$

$\implies \tt{v = \frac{ \cancel{400} \times 2}{ \cancel{20}}}$

$\implies \tt{v = 40 \: mps}$

so, the final velocity = 40 m/s

hence, we need to find the acceleration:

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

$\implies \tt{a = \frac{40 - 0}{20}}$

$\implies \tt{ a = \frac{ \cancel{40}}{ \cancel{20}}}$

$\red{ \underline{ \boxed{ \tt{ \therefore \: a = 2 \: m {s}^{ - 2}}}}}$

Hence, acceleration = 2m/s²

So, we Know that :

1 tonne = 1000 kg

so, 10 tonnes = 10×1000 kg

or, 10 tonnes = 10,000 kg

So, we Know the mass and acceleration

$\implies \tt{force = m \times a}$

$\implies \tt{force = 2 \times 10000}$

$\green{ \underline{ \boxed{ \tt{ \therefore \: force = 20000 \: n}}}}$

Hence, Force = 20000 N

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Additional Information:

Other Formula related to dynamics and motion :

1. v = u + at

2. S = ut + ½ a²

3. 2as = v² - u²

4. S = vt – ½ at 2

5. S = ½ (u + v) t

1. Force = mass×acceleration

2. Pressure = force/area

3. Impulse = force×time

4. momentum = mass×velocity.