Answer :
Final answer:
The car will take about 4.04 seconds to stop and it will travel about 89.36 meters in this time period while braking.
Explanation:
The subject of this question lies in the physics domain, specifically dealing with concepts of kinematics. To answer part a) we need to note that acceleration (deceleration in this case) is change in velocity divided by time. We must first convert the initial speed from km/h to m/s by dividing it by 3.6: 160 / 3.6 = 44.4 m/s. Then, we can use the acceleration to find the time: 44.4 / 11.0 = 4.04 s.
For question part b), we can use the formula for distance travelled under constant acceleration: D = vt + 0.5at2. Substituting the values, the distance travelled when braking is D = 44.4(4.04) - 0.5(11.0)(4.042) = 89.36 m.
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Answer:
a) t = 4.04 s
, b) x = 89.77 m
Explanation:
This is a one-dimensional kinematics exercise.
a) Let's use caution to find the time, the vehicle speed when stopped is zero
v = v₀ - at
let's reduce the magnitudes to the SI system
v₀ = 160 km /h (1000m / 1km) (1h / 3600s) = 44.44 m / s
0 = v₀ - at
t = v₀ / a
t = 44.44 / 11
t = 4.04 s
b) to calculate the distance traveled we use
x = v₀t - ½ a t²
x = 44.44 4.04 - ½ 11.0 4.04²
x = 179.54 - 89.77
x = 89.77 m
