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 metric tonnes

Hint - 1 metric tonne = 1000 kg. ​

Answer 1

Given :

• Initial Velocity (u) = 0

• Distance travelled (s) = 400 m

• Time (t) = 20 s

• Mass = 7 metric tonnes = 7000 kg

To Find :

• Acceleration of the truck

• Force acting on the truck

Solution :

$\dag\: \sf \: Calculation \: of \: acceleration$

From 2nd equation of motion :

★ s = ut + ½ at²

→ 400 = (0)(20) + ½ × a × (20)²

→ 400 = 0 + ½ × a × 400

→ 400 = a × 200

→ 400 = 200a

→ a = $\sf\cancel\dfrac{400}{200}$

→ a = 2 m/s²

$\therefore$ Acceleration of the truck is 2 m/s²

$\dag\: \sf \: Calculation \: of \: force$

From Newton's 2nd law of motion :

★ Force = Mass × Acceleration

→ Force = 7000 × 2

→ Force = 14000 N

$\therefore$ Force acting on the truck is 14000 newtons.

Answer 2

★ Given :

• Initial Velocity (u) = 0
• Distance travelled (s) = 400 m
• Time (t) = 20 s
• Mass = 7000 kg

★ To Find :

We have to find acceleration and force

★ Explanation :

Firstly, we will find the acceleration by using second equation of the motion.

$\large{\star{\underline{\boxed{\sf{s = ut + \dfrac{1}{2} at^2}}}}}$

$\sf{\dashrightarrow 400 = 0(20) + \dfrac{1}{2} \times a \times (20)^2} \\ \\ \sf{\dashrightarrow 400 = \dfrac{1}{2} \times a \times 400} \\ \\ \sf{\dashrightarrow 400 = 200a} \\ \\ \sf{\dashrightarrow a = \dfrac{400}{200}} \\ \\ \sf{\dashrightarrow a = 2} \\ \\ \large{\star{\underline{\boxed{\sf{a = 2 \: ms^{-2}}}}}}$

$\rule{200}{2}$

Now,

We will calculate Force

$\large{\star{\underline{\boxed{\sf{Force = Mass \times Acceleration}}}}} \\ \\ \sf{\dashrightarrow Force = 7000 \times 2} \\ \\ \sf{\dashrightarrow Force = 14000} \\ \\ \large{\star{\underline{\boxed{\sf{Force = 14,000 \: N}}}}}$

$\rule{400}{4}$