a body moving with a velocity of 50m/s is brought to rest in 30 secs by a constant retarding force. calculate the distance covered by the body?
Johnpaulvictor
26 Jan, 2025
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To find the acceleration (or deceleration) of the body, we can use the formula:
a = Δv / Δt
where:
a = acceleration (in m/s²)
Δv = change in velocity (in m/s)
Δt = time taken for the change (in s)
Given values:
v₁ = 50 m/s (initial velocity)
v₂ = 0 m/s (final velocity, since it's brought to rest)
Δt = 30 s
Now, let's calculate the acceleration:
a = (v₂ - v₁) / Δt
= (0 - 50) / 30
= -50 / 30
= -1.67 m/s²
The negative sign indicates that the acceleration is opposite to the direction of the initial velocity, meaning the body is decelerating.
So, the body is decelerating at a rate of 1.67 m/s².
Initial velocity (\(u\)) = 50 m/s
Final velocity (\(v\)) = 0 m/s (since the body is brought to rest)
Time (\(t\)) = 30 s
The distance (\(s\)) can be calculated using the kinematic equation for average velocity:\(s=\frac{u+v}{2}t\)
Step 2: Substitute the values into the formula and solve
Substitute the given values into the formula to find the distance.
frac{50\ m/s}+0
{m/s}}{2}\times 30
{s} (s=\frac {50}{2}\times 30\{m}(s=25\times 30\text{\ m}\)\(s=750\{ m}
Answer: The distance covered by the body is 750