Question

A truck starts from rest and rolls down a hill with a constant acceleration .It travels a distance of 400m in 20s. Find its acceleration . Find the force acting on it if its mass is 7 metric tonnes.​

Answer 1

$\huge\star{\underline{\mathtt{\red{A}\pink{n}\green{s}\blue{w}\purple{e}\orange{r}}}}$

Given:-

u = 0m/s

t = 20s

s = 400m

According to the second equation of motion,

$s = ut + \frac{1}{2} {at}^{2}$

We have,

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

$400 = 200a$

$a = {2m/s}^{2}$

Now, force = mass × acceleration

F = 7000×2

= 14000N

Answer 2

$\orange{\bigstar}$ Answer $\green{\bigstar}$

Acceleration = 2 m/s²

Force = 14 kN

Given

→ A truck starts from rest and rolls down a hill with a constant acceleration

→ It travels a distance of 400 m in 20 s

To Find

→ Acceleration

→ Force when mass is 7 tonnes

Formula

As the acceleration is constant throughout the motion , so we can apply equation's of kinematics .

$\orange{\bigstar}\ \; \bf s=ut+\dfrac{1}{2}at^2\ \; \green{\bigstar}\\\\\orange{\bigstar}\ \; \bf F=ma\ \; \green{\bigstar}$

Solution

Initial velocity , u = 0 m/s

Start's from rest

Distance , s = 400 m

Time , t = 20 s

Acceleration , a = ? m/s²

Apply 2nd equation of motion ,

$\to \rm s=ut+\dfrac{1}{2}at^2\\\\\to \rm 400=(0)(20)+\dfrac{1}{2}a(20)^2\\\\\to \rm 400=0+200a\\\\\to \rm 200a=400\\\\\to \rm a=2\ m/s^2\ \pink{\bigstar}$

So , Acceleration = 2 m/s²

Now ,

Mass , m = 7 tonnes

⇒ m = 7000 kg  [ ∵ 1 tonne = 1000 kg ]

Apply Newton sir's Second Law ,

$\to \rm F=ma\\\\\to \rm F=(7000)(2)\\\\\to \rm F=14000\ N\\\\\to \rm F=14\ kN\ \red{\bigstar}$