A truck starts from rest and rolls down a fills with a constant acceleration . it travels a distance of 400 m in 20s . find its accelaration. find its accelaration on it if its mas is 7 tonnes.
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4
From the equation,
s = ut+(1/2)at²
u = 0
s= 400 m
t= 20 sec
⇒400 = (1/2) a×(20)²
⇒800 = 400 a
⇒a = 2 m/s²
1 tonne = 1000 kg
so, 7 tonnes = 7000 kg
F = ma = 7000×2 = 14000 N
s = ut+(1/2)at²
u = 0
s= 400 m
t= 20 sec
⇒400 = (1/2) a×(20)²
⇒800 = 400 a
⇒a = 2 m/s²
1 tonne = 1000 kg
so, 7 tonnes = 7000 kg
F = ma = 7000×2 = 14000 N
Answered by
2
Given,
u = 0
s= 400 m
t= 20 sec
As we know that,
s = ut+(1/2)at²
By implying values, we get
⇒400 = (1/2) a×(20)²
⇒800 = 400 a
⇒a = 2 m/s²
1 tonne = 1000 kg
so, 7 tonnes = 7000 kg
F = ma = 7000×2 = 14000 N
u = 0
s= 400 m
t= 20 sec
As we know that,
s = ut+(1/2)at²
By implying values, we get
⇒400 = (1/2) a×(20)²
⇒800 = 400 a
⇒a = 2 m/s²
1 tonne = 1000 kg
so, 7 tonnes = 7000 kg
F = ma = 7000×2 = 14000 N
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