Question

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

Answer 1

According to the question,

Initial velocity u = 0 (since the body starts from rest)
s, that is, the displacement = 400m

Finally, we are given the time = 20s


From the second equation of motion we have


s = ut + (1/2)at²

Substituting the value of s, u and t we get,

$400 = 0 \times 20 + \frac{1}{2} \times a \times {20}^{2} \\ \\ = > 400 = 0 + \frac{1}{2} \times a \times 400 \\ \\ = > 400 = 200a \\ \\ = > a = \frac{400}{200} \\ \\ = > a = 2$


So the acceleration (a) is 2m/s²


Now 1 Metric ton (tonne) = 1000kg

So 7 metric tons = 7 × 1000

=> 7 metric tons = 7000 kg

From Newton's second law we know that,

F = ma

Where F is force, m is mass(in kg) and a (in m/s²) is acceleration

=> F = 7000 × 2

=> F = 14000 kg m/s² or 14000 Newtons


So your answer

Acceleration = 2 m/s²
Force = 14000 N


Hope it helps dear friend ☺️✌️

Answer 2

It's the solution....

Given : u = 0, s = 400 m, t = 20 s, m = 7 metric ton = 7000 kg, a= ?,F = ?

Using second equation of motion

s = ut + 1/2 at_2

400 = 0* 20+ 1/2*a*(20)_2

400 = 0 + 1/2 * a * 400

400 = 200 a

400/200 = a

2m/s = a

F = ma

F= 7000* 2

F = 14000N