Question

1. Anand leaves his house at 8.30 am for his school. The school is 2km away and classes start at 9 am. If he walks at a speed of 2 km/hr for the first kilometre , at what speed should he walks the second kilometre to reach just in time?.

2. A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/hr in 10 minutes . Find its acceleration.

3. A scooter travelling at 10 m/s speeds up to 20 m/s in 4s . Find the acceleration of the scooter.

4. An object dropped from a cliff falls with a constant acceleration of 10 m/s2 . Find it's speed after 5s seconds it was dropped.

5. A driver of a car A travelling at 52 km/hr applies the brake and the car stops in 5 s . A driver of car B going at 3 km/hr applies brake and the car stops in 10 s. calculate the acceleration of car A and car B . Also find which car has more acceleration than other.

$\mathbb {\bold{PLEASE \: ANSWER \: THESE \: }} \\ \mathbb{ QUESTIONS \: WITH \: STEP \: } \\ \mathbb{ BY \: STEP \: EXPLANATION}$
$\tt{ \gray{take \: time = \frac{1}{2} h \:to \: complete }}$

$\tt{it \: is \: 90 \: pts \: do \: well}$
​

Answer 1

Answer:

Okey let's see from begining

Follow #me

Explanation:

1)A-----------B-----------C

let A is anand house

B is1km distance apart from his house

C is school

we know time =distance/speed

tAB=1/3 hour

tAB is time takenfrom A to B

tBC=1/2-1/3 =1/6hour

so, speed=distance/time=1/(1/6)=6km/h

2)Known Terms:-

u = Initial speed

v = Final speed

a = Acceleration

t = Time

Given:-

Initial speed of the train, u = 0 km (Because train starts from the rest.

Final speed of the train, v = 40 km/h = × 40 = 11.11 m/s.

Time taken to attain the speed = 10 minutes = 10 × 60 = 600 sec.

To Calculate:-

Acceleration of the train.

Formula to be used:-

⇒

Solution:-

Putting all the values, we get

⇒

⇒

⇒

⇒

Hence, Acceleration of the train is 0.0185 m/s².

3)Hey there ,

Praneeth from brainly.in team ready to help !

A bit correction in the question

Question :- a scooter travelling at 10m/s speed up to 20m/s in 4 seconds .find the acceleration of Scooter

Answer :-

Intial velocity =u=10m/s

Final velocity =v=20 m/s

Time=t=4s

Acceleration of the scooter =v-u/t

=20-10/4

=10/4

=2.5 m/s²

The acceleration produced is 2.5m/s²

4)Acceleration =

Solution:

The initial velocity is zero as it's a free fall.  

Thereby, applying "equation of motion", we can calculate the velocity after 2 seconds of fall.  

The "first equation of motion" is to be applied that is, v = u + a t  

v = 0 + a t  

v = 10 × 2  

v = 20 m/s

5)Given : 

First Car  A: 

Initial Velocity= u= 52 km/hr=52x5/18=14.4 m/s

Final velocity= V= 0m/s

time =t=5sec

Refer attachment for Graph :

Distance travelled : Area of traingle AOB =1/2 OB x AO

=(1/2)x 14.4 x 5=72/2=36m

Second Car  B:

Initial Velocity= u= 3 km/hr=3x5/18=0.83m/s

Final velocity= V= 0m/s

time =t=10sec

Refer attachment for Graph :

Distance travelled : Area of traingle COD =1/2 OD x CO=(1/2)x 0.83 x 10=4.1m

So, after applying of brakes , Car A has travelled a distance of 3m and car B has travelled a distance of 4.1m.

∴ Car B has travelled farther than Car A after the brakes are applied.

Hopes

Answer 2

$\textit{\large{\underline{\underline{EXPLANATION:}}}}$

Sorry for 1st question. I'm not sure about it

$\rule{200}2$

$\textbf{\underline{\underline{Que 2,,}}}</p><p>$

$\sf{Given}\begin{cases}\sf{u=0kmh^{-1}}\\ \sf{v=40kmh^{-1}}\\ \sf{t=10min=\frac{1}{6}h}\sf{a=??}\end{cases}$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{v=u+at}}}$

$\sf{\implies{40=10a}}$

$\sf{\implies{\frac{40}{10}=a}}$

$\sf{\implies{4=a}}$

Therefore,,

$\sf{\boxed{\red{a=4kmh^{-2}}}}$

$\rule{200}2$

$\textbf{\underline{\underline{Que 3,,}}}</p><p>$

$\sf{Given}\begin{cases}\sf{u=10ms^{-1}}\\ \sf{v=20ms^{-1}}\\ \sf{t=4s}\sf{a=??}\end{cases}$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{v=u+at}}}$

$\sf{\implies{20=10+4a}}$

$\sf{\implies{\frac{20-10}{4}=a}}$

$\sf{\implies{2.5=a}}$

Therefore,,

$\sf{\boxed{\red{a=2.5ms^{-2}}}}$

$\rule{200}2$

$\textbf{\underline{\underline{Que 4,,}}}</p><p>$

$\sf{Given}\begin{cases}\sf{u=0ms^{-1}}\\ \sf{v=-??}\\ \sf{t=5s}\sf{a=-10ms^{-2}}\end{cases}$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{v=u+at}}}$

$\sf{\implies{v=0+(-10)(5)}}$

$\sf{\implies{v=-50}}$

Therefore,,

$\sf{\boxed{\red{v=-50ms^{-1}}}}$

$\rule{200}2$

$\textbf{\underline{\underline{Que 5,,}}}</p><p>$

$\textbf{\underline{For car A,,}}</p><p>$

$\sf{Given}\begin{cases}\sf{u=52kmh^{-1}=\frac{26}{9}ms^{-1}}\\ \sf{a=??}\\ \sf{t=5s}\sf{v=0ms^{-1}}\end{cases}$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{v=u+at}}}$

$\sf{\implies{0=\frac{26}{9}+5a}}$

$\sf{\implies{\frac{-26}{9\times{5}}=a}}$

$\sf{\implies{\frac{-26}{45}=a}}$

Therefore,,

$\sf{\boxed{\red{a={-26}{45}ms^{-2}}}}$

$\textbf{\underline{For car B,,}}</p><p>$

$\sf{Given}\begin{cases}\sf{u=3kmh^{-1}=\frac{5}{6}ms^{-1}}\\ \sf{a=??}\\ \sf{t=10s}\sf{v=0ms^{-1}}\end{cases}$

$\textbf{\underline{Using equation of motion,,}}</p><p>$

$\mathtt{\boxed{\blue{v=u+at}}}$

$\sf{\implies{0=\frac{5}{6}+10a}}$

$\sf{\implies{\frac{-5}{6\times{10}}=a}}$

$\sf{\implies{\frac{-1}{12}=a}}$

Therefore,,

$\sf{\boxed{\red{a={-1}{12}ms^{-2}}}}$

Now,,

$\sf{\boxed{\red{a={-26}{45}=-0.57}}}$

$\sf{\boxed{\red{a={-1}{12}=-0.83}}}$

$\rule{200}2$

Hope it helps ❣️❣️❣️