Question

A truck start from rest and rolles down on hill with constant acceleration.It travels a distance of 400 m in 20 s.find the acceleration. find the force acting on it if it's much is 7 metric tonne.​

Answer 1

Explanation:

Initial velocity =u = 0

Distance =s = 400 m

time = t= 20 sec

a = acceleration

Equation of motion :-

s = ut + 1/2at²

400 = 0 + 1/2 × a × 20 × 20

400 = 200 a

a = 2

acceleration = a = 2 m/s²

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Force = F

Mass = m = 7000 kg

Acceleration = a = 2 m/s²

F = ma

= 7000 × 2

= 14000 N

Force acting on it is 14000 N

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Answer 2

Answer:

Acceleration=2 m/s^2

Force acting on it=14000 N

Given:-

U=0,t=20 s and S=400m

By using 2nd equation of Motion

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

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

$400 = 0 + \frac{1}{2}a \times 400$

$400 = 0 + 1 \times a \times 200$

$400 = a \times 200$

$\frac{400}{200} = a$

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

We know that

Force=ma

mass=7 metric tonne

1 metric tonne=1000kg

7 metric tonne=7000kg

_________

Force=7000×2

=14000 N

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Acceleration=2m/s^2 and Force=14000 N