physics numericals
motion in one dimension
1.if a sports car can go from rest to 27 ms-^1 in 9.0 seconds what is the magnitude of its average accelaration
Answers
Answered by
1
a = dv/dt
a = 27 - 0/9
a = 3 m/s^2
a = 27 - 0/9
a = 3 m/s^2
Answered by
4
Given:
u=0 (body is at rest initially)
v=27 m/s
time interval = 9 s
a = ∆v/t
a(0,4) = (v4-v0)/(t4-0)
= (27-0)/9
= 3 m/s²
There is an average acceleration of 3 metres per second squared for the time interval (0,4)
u=0 (body is at rest initially)
v=27 m/s
time interval = 9 s
a = ∆v/t
a(0,4) = (v4-v0)/(t4-0)
= (27-0)/9
= 3 m/s²
There is an average acceleration of 3 metres per second squared for the time interval (0,4)
Similar questions