2.6 Consider a car that is initially traveling along a straight stretch of highw
ID: 1562217 • Letter: 2
Question
2.6
Consider a car that is initially traveling along a straight stretch of highway at 15 m/s. At t = 0 the car begins to accelerate at 2.0 m/s^2 in order to pass a truck. The final velocity is upsilon_x = 25 m/s. Calculate how far the car travels during its 5.0 s of acceleration. SET UP (Figure 1) shows what we draw. Note that upsilon__0 x = +15 m/s and a_x = +2.0 m/s^2. SOLVE We want to solve for x - x_0. the distance traveled by the car during the 5.0 s time interval. The acceleration is constant, so we can use the following equation to find x - x_0 = upsilon_0 x t+ 1/2 a_xt^2 = (15 m/s)(5.0 s)+ 1/2 (2.0 m/s^2)(5.0 s)^2 = 75 m+ 25 m = 100 m REFLECT If the speed were constant and equal to the initial value upsilon_0 = 15 m/s. the car would travel 75 m in 5.0 s. It actually travels farther because the speed is increasing. The final velocity is upsilon_x = 25 m/s, so the average velocity for the 5.0 s segment of motion is upsilon_av, x = upsilon_0 z + upsilon_x/2 = 15 m/s + 25 m/s/2 = 20 m/s An alternative way to obtain the distance traveled is to multiply the average velocity by the time interval. When we do this, we get x - x_0 = upsilon_av, t = (20 m/s)(5.0 s) = 100 m. If the car maintains its constant acceleration for a total time of 30 s, what total distance does it travel? Express your answer in meters to the nearest integer.Explanation / Answer
given the car is travelling at constant acceleration a =2m/sec^2
initial vel =u = 15m/s
time = 30sec
total distance travelled s = ut+1/2*a*t^2
s = 15*30 + 1/2 *2 *900 = 900+450 = 1350m
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.