Question

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.)​

Answer 1

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.

Analysis :-

Here, we are given with distance, time and initial velocity of the truck.

Firstly, using the second equation of motion find the acceleration accordingly.

Substitute the values with you in the formula of force and you can get the value easily.

Solution :-

We know that,

• u = Initial velocity
• a = Acceleration
• f = Force
• v = Final velocity
• s = Distance
• t = Time

Using the formula,

$\underline{\boxed{\sf Second \ equation \ of \ motion=s=ut+\dfrac{1}{2} at^2}}$

Given that,

Distance (s) = 400 m

Time taken (t) = 20 s

Initial velocity (u) = 0 m/s

Substituting their values,

400 = 0(20) + 1/2 (a) 400²

a = 2 m/s⁻²

Therefore, the acceleration is 2 m/s⁻².

Using the formula,

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

Given that,

Mass (m) = 7 tonnes = 7000 kg

Acceleration (a) = 2 m/s⁻²

Substituting their values,

f = 7000 × 2

f = 14000 N

Therefore, the force acting on it is 14000 N.

Answer 2

$\huge {\blue {\underline {\underline {\mathtt{Question}}}}}$

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.)

$\huge {\pink {\underline {\underline {\mathtt{Answer }}}}}$

$\huge {\purple {\mathfrak{Given :}}}$

• Initial velocity u = 0 m/s
• distance s = 400m
• Time t = 20 s
• mass m = 7000kg

$\huge {\orange{\mathfrak{To \:find :}}}$

• Acceleration
• Force

$\huge {\red {\mathfrak{Solution :}}}$

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

$\sf:\implies 400 = 0 × 20 + \dfrac {1}{2}a × 20 × 20$

$\sf:\implies 400 = \dfrac {1}{2}a × 400$

$\sf:\implies 400 = 200a$

$\sf:\implies a = 2 m/s^2$

${\pink {\boxed {\sf{\underline {Acceleration = 2 m/s^2}}}}}$

________________________________________________

${\blue {\boxed {\sf{Force = m × a }}}}$

$\sf:\implies F = 7000 × 2$

$\sf:\implies F = 14 , 000 N$

${\orange{\boxed {\underline{\sf{Force = 14000 N }}}}}$

_____________________________________

${\boxed {\underline {\sf{☆Remember : ☆}}}}$

$\sf v = u + atv=u+at$

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

${\sf {v^2 = u^2 + 2as}}$