A particle starts from origin t=0 with a velocity 5.0 ^i m/s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0 ^i + 2.0^j) m/s squ.
a) What is the y coordinate of the particle at the instant its x coordinate is 84m?
b)What is the speed of the particle at this time?
Please help! :o
Answers
Hi there...here's ur answer..
The y-coordinate of the particle is 36 m at the instant its x-coordinate is 84m and the speed of the particle at this time is 25.94 m/s.
Given : A particle starts from origin t=0 with a velocity 5.0î m/s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0î + 2.0ĵ) m/s².
To find : The the y-coordinate of the particle at the instant its x-coordinate is 84m. And, the speed of the particle at this time.
Solution :
We can simply solve this numerical problem, by using the following process. (our goal is to calculate the required y-coordinate and the required speed)
So,
Velocity = 5î m/s
Acceleration = (3.0î + 2.0ĵ) m/s²
Now, according to kinematics :
s = ut + ½ at²
Towards x-axis of x-y plane (i.e. uₓ = 5 m/s , aₓ = 3 m/s²) :
x = uₓ + ½ aₓt
x = (5 × t) + (½ × 3 × t²)
x = 5t + (³/₂ × t²)
84 = 5t + (³/₂ × t²) [as, x-coordinate = 84]
(3t²/2) + 5t = 84
(3t²/2) + 5t - 84 = 0
(3t²+10t-168)/2 = 0
3t²+10t-168 = 0
3t²+28t-18t-168 = 0
t(3t+28) - 6(3t+28) = 0
(t-6) (3t+28) = 0
Either,
(t-6) = 0
t = 6 s
Or,
(3t+28) = 0
3t = -28
t = -9.33 s
Now, time can't be negative, that's why we will omit (3t+28) = 0
So, t = 6 s
Towards y-axis of x-y plane (i.e. uᵧ = 0 m/s , aᵧ = 2 m/s², t = 6 s) :
y = uᵧt + ½ aᵧt²
y = (0 × 6) + (½ × 2 × 6²)
y = 36m
Now,
- vₓ = uₓ + aₓt = [5 + (3×6)] = 23 m/s
- vᵧ ,= uᵧ + aᵧt = [0 + (2×6)] = 12 m/s
Finally,
The speed of the particle :
= √[(vₓ)² + (vᵧ)²]
= √[(23)²+(12)²]
= √(529+144)
= √673
= 25.94 m/s
Hence, The y-coordinate of the particle is 36m at the instant its x-coordinate is 84m and the speed of the particle at this time is 25.94 m/s.
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