Question

A truck starts from the 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 6 tonnes ( hint: 1 tonne = 1000 kg​

Answer 1

Given

u = 0

t = 20 s

s = 400 m

a = ?

s = ut + at²/2

400 = 0*20 + a*20²/2

400 = 400a/2

400 = 200a

a = 2m/s²

m = 6*1000 = 6000 kg

a = 2 m/s²

F = ma

F = 6000*2

F = 12000 N

Answer 2

Answer:

Acceleration,a = 2ms^-2

Force,f = 1400N

Explanation:

Given :

Initial velocity of the truck,u => 0 [Since truck starts from rest]

Distance travelled,s => 400m

Time taken,t => 20s

Mass,m => 7 tonnes = (7 × 1000)kg = 7000kg

To Find :

Accelaration,a => ?

Force,f => ?

Solution :

We know that :-

\rm{s=ur+\dfrac{1}{2}at^2

\rightarrow\rm{400m=0\times 20s+\dfrac{1}{2}\times a\times(20)s^2

\rightarrow \rm{400m=0+\dfrac{1}{2}\times a\times400s^{2}

\rightarrow\rm{}400m=\dfrac{1}{1}\times a\times 200s^2→400m=

1

1

×a×200s

2

\rightarrow\rm{400m=a\times200s^2}→400m=a×200s

2

\rightarrow \rm{a=\dfrac{400m}{200s^2}

\rightarrow\rm{a=2ms^{-2}}→a=2ms

−2

Force = m × a

⇒ F = 7000kg × 2ms^-2

⇒ F = 1400N