A car travelling at a speed of 10 m/s is brought to rest in 20 seconds by applying brakes. Calculate the acceleration and distance travelled during this time
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8
u = 10 m/s
v =0
t = 20 sec
From 1st equation of motion,
v =u+at
⇒0 =10 +20a
⇒20a = - 10
⇒a = -0.5 m/s²
now, from 3rd equation,
v²-u² = 2as
⇒(0)²-(20)² = 2×(-0.5)×s
⇒s = 400 m
v =0
t = 20 sec
From 1st equation of motion,
v =u+at
⇒0 =10 +20a
⇒20a = - 10
⇒a = -0.5 m/s²
now, from 3rd equation,
v²-u² = 2as
⇒(0)²-(20)² = 2×(-0.5)×s
⇒s = 400 m
Answered by
3
Given,
Initial Velocity = 10 m/s
Final Velocity =0
Time taken= 20 sec
According to1st equation of motion,
v =u+at
By implying values,
⇒0 =10 +20a
⇒20a = - 10
⇒a = -0.5 m/s²
According to 3rd equation of motion,
v²-u² = 2as
By implying values
⇒(0)²-(20)² = 2×(-0.5)×s
⇒s = 400 m
Initial Velocity = 10 m/s
Final Velocity =0
Time taken= 20 sec
According to1st equation of motion,
v =u+at
By implying values,
⇒0 =10 +20a
⇒20a = - 10
⇒a = -0.5 m/s²
According to 3rd equation of motion,
v²-u² = 2as
By implying values
⇒(0)²-(20)² = 2×(-0.5)×s
⇒s = 400 m
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