Question

A firetruck is also on its way to the emergency. The firetruck accelerates at 0.75 m/s² for 37 seconds from rest. How far did it travel in that time?

Answer 1

The firetruck traveled a distance of 513.375 meters in 37 seconds.

Step-by-step explanation:

To find the distance traveled by the firetruck, we can use the equation d = v0t + 0.5at2, where d is the distance, v0 is the initial velocity, a is the acceleration, and t is the time.

In this case, the firetruck starts from rest, so the initial velocity is 0. The acceleration is given as 0.75 m/s2, and the time is 37 seconds.

Plugging in these values into the equation, we have d = (0)(37) + 0.5(0.75)(37)2 = 0 + 0.5(0.75)(1369) = 0 + 0.5(1026.75) = 513.375 meters.

Therefore, the firetruck traveled a distance of 513.375 meters in 37 seconds.

Answer 2

The firetruck travels 513.375 meters in 37 seconds when it accelerates from rest at a rate of 0.75 m/s².

Step-by-step explanation:

The student asks how far a firetruck travels in 37 seconds when it accelerates from rest at a rate of 0.75 m/s². To calculate the distance traveled, we can use the kinematic equation:

s = ut + ½ at²

where:

• s is the distance
• u is the initial velocity (0 m/s since it's from rest)
• a is the acceleration (0.75 m/s²)
• t is the time (37 s)

Plugging in the values:

s = (0) · 37 + ½ · 0.75 · (37)² = 0 + 0.375 · 1369 = 513.375 meters