Question

5. 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
7 tonnes (Hint: 1 tonne = 1000 kg.)


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Answer 1

Solution :-

Given :-

Distance, S = 400m

Time taken, t = 20s

Initial velocity, u  = 0

We can find the acceleration of the truck by the third equation of kinematics.

$\boxed{\bf S= ut + \dfrac{1 } {2} at^2}$

Substitute the given values :-

$\longrightarrow \sf 400 = 0\times 20 + \dfrac{1}{2} \times a \times 20 \times 20 \\\\\longrightarrow 400 = 0 + a \times 10 \times 20 \\\\\longrightarrow 400 = 200a \\\\\longrightarrow \bf a = 2$

Acceleration of the truck = 2 m/s²

We know :-

F = ma

m = 7 tonnes

   = 7 × 1000 kg

$\longrightarrow \sf F= 7 \times 100 0 \times 2 \\\\\longrightarrow\bf F = 14000N$

Force acting on truck = 14000N

Answer 2

◇Given :-

• Distance covered by the truck = 400 m

• Time taken to cover the distance 20 s Initial velocity of the truck = 0

◇To Find :-

• The acceleration.

• The force acting on it.

◇Solution:-

• u = Initial velocity

• a = Acceleration

• f = Force

• v = Final velocity

• s = Distance

• t = Time

Using Formula,

$\\ { \underline{ \boxed{ \sf{second \: equation \: of \: motion \: = \:s\:= \: ut + \frac{1}{2}a {t}^{2} }}}}$

• Distance (s) = 400 m

• Time taken (t) = 20 s

• Initial velocity (u) = 0 m/s =

Substituting their values,

$\\ \sf \: 400 = 0(20) + \frac{1}{2} (a) \: {400}^{2}$

$\sf \: a = 2 m/s \: </p><p>$

Therefore, the acceleration is 2 m/s.

Using formula,

$\\ { \underline{ \boxed{ \sf{Force = Mass \times Acceleration}}}}$

Given that,

• Mass (m) = 7 tonnes = $\sf{7000\: kg}$

• Acceleration (a) =$\sf}2 \:m/s}$

Substituting their values,

• $\sf{f\: =\: 7000\: \times 2}$

• $\sf{f \:= \:14000 \:N}$

Therefore, the force acting on it is

14000 N.