3) A truck is traveling at 35 m/s. Because of an obstruction on the road, it is forced to stop.
a) If the truck can decelerate at a constant 5.0 m/s2 how long will it take to stop?
b) How far does the truck travel before it stops?
To find the answers to these questions, we can use the equations of motion and the concept of uniform deceleration.
a) The first question asks for the time it takes for the truck to stop. We can use the equation of motion:
v = u + at,
where:
- v is the final velocity (which in this case is 0, as the truck will come to a stop),
- u is the initial velocity (which is 35 m/s),
- a is the acceleration (which is the deceleration rate, -5.0 m/s² in this case), and
- t is the time.
Rearranging the equation to solve for time (t), we have:
t = (v - u) / a.
Substituting the given values, we get:
t = (0 - 35) / -5.0 = 7 seconds.
Therefore, it will take the truck 7 seconds to come to a stop.
b) The second question asks for the distance the truck travels before stopping. We can use the equation of motion:
s = ut + (1/2)at²,
where:
- s is the displacement (which is the distance the truck travels before stopping),
- u is the initial velocity (35 m/s),
- t is the time (7 seconds), and
- a is the acceleration (deceleration rate, -5.0 m/s²).
Substituting the given values, we have:
s = (35)(7) + (1/2)(-5.0)(7)²
= 245 + (-5.0)(49)
= 245 - 245
= 0.
Therefore, the truck travels a distance of 0 meters before coming to a stop.
To summarize:
a) It takes 7 seconds for the truck to stop.
b) The truck travels a distance of 0 meters before stopping.