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

Given :-

Distance covered by the truck, s = 400m

Time taken, t = 20s

Initial velocity of the truck, u = 0m/s

To find :-

(i) accerlation of the truck

(ii) froce acting on it

Solution :-

(i) as we are providing with distance (s), initial Velocity (u) and Time we can use 2nd equation of motion

so, using 2nd equation of motion .i.e.,

➠ s = ut + 1/2at²

➠ 400 = (0)(20) + 1/2(a)(20)²

➠ 400 = 0 + 200a

➠ 400 = 200a

➠ a = 400/200

➠ a = 2m/s²

thus, the acceleration of the truck is 2 m/s²

(ii) using Newton's 2nd law of motion .i.e.,

The rate of change of linear Momentum of a body with time is proportional to the net external force acting on it

ln other words,

• f = ma

1st let's convert tonnes into kg

1 ton = 1000kg

7 tonnes = 7 × 1000 = 7000kg

According to Question,

• mass, m = 7tonnes .i.e., 7000kg
• accerlation, a = 2m/s²

➠ f = ma

➠ f = (7000)(2)

➠ f = 14000N

thus, the force acting on the truck is 14000N

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Answer 2

Answer:

Given :-

Distance covered by the truck, s = 400m

Time taken, t = 20s

Initial velocity of the truck, u = 0m/s

To find :-

(i) accerlation of the truck

(ii) froce acting on it

Solution :-

(i) as we are providing with distance (s), initial Velocity (u) and Time we can use 2nd equation of motion

so, using 2nd equation of motion .i.e.,

➠ s = ut + 1/2at²

➠ 400 = (0)(20) + 1/2(a)(20)²

➠ 400 = 0 + 200a

➠ 400 = 200a

➠ a = 400/200

➠ a = 2m/s²

thus, the acceleration of the truck is 2 m/s²

(ii) using Newton's 2nd law of motion .i.e.,

The rate of change of linear Momentum of a body with time is proportional to the net external force acting on it

ln other words,

f = ma

1st let's convert tonnes into kg

1 ton = 1000kg

7 tonnes = 7 × 1000 = 7000kg

According to Question,

mass, m = 7tonnes .i.e., 7000kg

accerlation, a = 2m/s²

➠ f = ma

➠ f = (7000)(2)

➠ f = 14000N

thus, the force acting on the truck is 14000N