Question

A truck starts from rest with a constant acceleration.it travels a distance of 400m in 20 sec.find the speed accessed by the truck

Answer 1

Answer:

40 m/s

Explanation:

Given :

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

To find :

• Speed accessed by the truck

Using second equation of motion :

S=ut+½at²

400=0×20+½×a×20²

400=0+½×400×a

400=200a

a=400/200

a=2m/s²

The acceleration of the truck js equal to 2 m/s²

Now using the first equation of motion :

V=u+at

V=0+2×20

V=0+40

V=40 m/s

The final speed acquired by the truck is 40 m/s

Answer 2

✯✯ QUESTION ✯✯

A truck starts from rest with a constant acceleration.it travels a distance of 400m in 20 sec.find the speed accessed by the truck ..

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✰✰ ANSWER ✰✰

$\longmapsto\tt{Distance(s)=400m.}$

$\longmapsto\tt{Time\:Taken(t)=20.}$

$\longmapsto\tt{Initial\:Velocity(u)=0}$

$\longmapsto\tt{Acceleration(a)=?}$

$\longmapsto\tt{Speed(v)=?}$

➥Now ,

Firstly we will find the acceleration of truck...

➥Using 2nd Equation of Motion : -

$\longmapsto\tt{\small{\boxed{\bold{\bold{\red{\sf{s=ut+\dfrac{1}{2}{at}^{2}}}}}}}}$

➥Putting Values : -

$\longmapsto\tt{400=(0)\times{(20)}+\dfrac{1}{2}\times{a}\times{{(20)}^{2}}}$

$\longmapsto\tt{400=0+\dfrac{1}{\cancel{2}}\times{a}\times{{\cancel{400}}}}$

$\longmapsto\tt{400=200a}$

$\longmapsto\tt{a=\cancel\dfrac{400}{200}}$

$\red\longmapsto\:\large\underline{\boxed{\bf\green{a}\orange{=}\purple{2}}}$

☛Acceration of Truck is 2m/s²...

_______________________

Now , we will find the speed of truck ....

➥Using 1st Equation of Motion : -

$\longmapsto\tt{\small{\boxed{\bold{\bold{\purple{\sf{v=u+at}}}}}}}$

➥Putting Values : -

$\longmapsto\tt{v=(0)+2\times{20}}$

$\longmapsto\tt{v=2\times{20}}$

$\orange\longmapsto\:\large\underline{\boxed{\bf\red{v}\blue{=}\pink{40}}}$

☛Speed of truck is 40m/s..