Answer:
s = 23.34 m
Explanation:
given,
F = 16.8 t
mass = 45 Kg
time = 5 s
distance traveled by the object = ?
now,
Force acting at time 5 s
F = 16.8 x 5
F = 84 N
we know,
F = ma
84 = 45 x a
a = 1.867 m/s²
initial velocity = 0 m/s
using equation of motion for velocity calculation
v = u + at
v = 0 + 1.867 x 5
v = 9.335 m/s
again using equation of motion for distance calculation
v² = u² + 2 a s
9.335² = 0 + 2 x 1.867 x s
s = 23.34 m
distance traveled is equal to 23.34 m.
When A mysterious rocket-propelled object of mass 45.0 kg is initially at rest in the middle of the horizontal, friction-less surface of an ice covered lake. Then a force directed east and with magnitude F(t) = (16.8 N/s) t is applied, The object will travel 22m in the first 5.00 s after a force is applied.
Force (F ) = mass (m)× Acceleration(a)
F = ma.
a = F/m............... Equation 1
From the question,
First we need to find the value of the acceleration
Given: m = 45.0 kg, F = 16.8t, where t = time. given t = 5.00 s
subtitute these values into equation 2
Therefore,
a = 16.8(5)/45
a = 1.76 m/s²
Secondly, Using the equation of motion,
s = ut+ at²/2................. Equation 2
Where, u = initial velocity, u = initial velocity, a = acceleration, s = distance.
Given: u = 0 m/s ( at rest), a = 1.76 m/s², t = 5 s
Substitite this values into equation 2
s = 0(5) + 1.76(5²)/2
s = 22 m
Hence, the object will travel 22 m in the first 5.00 seconds
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