A boat initially traveling at 10m/s accelerates uniformly at the rate if 5 m/s^2 for 10 seconds. How far does the boat travel during this time?
The correct answer and explanation is :
Answer:
To find the distance traveled by the boat, we use the equation of motion:
$$
s = ut + \frac{1}{2}at^2
$$
Where:
- $s$ = distance traveled (in meters)
- $u$ = initial velocity = 10 m/s
- $a$ = acceleration = 5 m/s²
- $t$ = time = 10 seconds
$$
s = (10)(10) + \frac{1}{2}(5)(10^2) \
s = 100 + \frac{1}{2}(5)(100) \
s = 100 + 250 = \boxed{350 \text{ meters}}
$$
Explanation (Approx. 300 words):
To determine how far a boat travels while accelerating, we apply the kinematic equations of motion, which describe the relationships between velocity, acceleration, time, and displacement under uniform acceleration.
In this scenario, the boat starts with an initial speed of 10 meters per second (m/s). It then undergoes a constant acceleration of 5 meters per second squared (m/s²) over a time span of 10 seconds. Because the acceleration is uniform, we can safely use the equation:
$$
s = ut + \frac{1}{2}at^2
$$
This formula combines the distance traveled at the initial speed with the additional distance due to acceleration. The term $ut$ calculates how far the boat would go in 10 seconds if it didn’t accelerate (i.e., $10 \times 10 = 100$ meters). The second part, $\frac{1}{2}at^2$, accounts for the extra distance covered as the boat speeds up — in this case, $\frac{1}{2} \times 5 \times 100 = 250$ meters.
Adding both gives the total distance:
$$
100 + 250 = 350 \text{ meters}
$$
This means the boat covers 350 meters during the 10-second period. This example highlights how motion under constant acceleration accumulates distance more rapidly over time. The effect of acceleration becomes especially noticeable the longer it continues, as the squared time term causes distance to grow faster than with constant speed alone. Understanding this concept is key in physics, especially in mechanics and motion-related problems.