A car moving with constant acceleration covered the distance between two points 52.4 m apart in 5.96 s. Its speed as it passes the second point was 14.6 m/s. (a) At what prior distance from the first point was the car at rest?

Question

Benjamin Asiedu · Accepted Answer

The car was at rest #"2.29 m"# before the first point.

Here, the idea is that the car will accelerate steadily once it gets going. This implies that you can write using the acceleration and the speed at which it passes the first point. #v_1^2 = underbrace(v_0^2)_(color(blue)(=0)) + 2 * a * d" "#, where #v_1# - the speed it has at the moment it passes the first point; #v_0# - the initial velocity, equal to zero since the car's starting from rest; #d# the disance before it passes the first point. The values you have for #v_1# and #a# are correct, but I'll show how to get them for other students interested in how this problem can be solved. You basically work with two equations with two unknowns, #v_1# and #a#. To begin with, you are aware of #s = v_1 * t + 1/2 * a * t^2" "#, where #s# - the distance between the two points; #t# - the time needed to go from the first point to the second point; Given that you are aware of the car's speed at the second point, you can write #v_2 = v_1 + a * t implies v_1 = v_2 - a * t# Re-enter this into the equation above to obtain #s = (v_2 - at) * t + 1/2 * a * t^2# Enter your values to get; for simplicity, I'll omit the units. #52.4 = (14.6 - 5.96 * a) * 5.96 + 1/2 * a * 5.96""^2# This will ultimately result in you #52.4 = 87.016 - 35.52a + 17.76a# #17.76a = 34.616 implies a = 34.616/17.76 = "1.95 m/s"""^2# When the vehicle crosses the first point, its speed will be #v_1 = 14.6 - 1.95 * 5.96 = "2.99 m/s"# Now, you know that the car started from rest and reached a speed of #"2.99 m/s"# by the time it passed the first point. This means that you have #v_1^2 = 2 * a * d implies d = v_1^2/(2 * a) = (2.99^2"m"^color(red)(cancel(color(black)(2)))/color(red)(cancel(color(black)("s"^2))))/(2 * 1.95color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2)))) = color(green)("2.29 m")# As an alternative, you can first determine how long it took the vehicle to arrive at the starting point. #v_1 = underbrace(v_0)_(color(blue)(=0)) + a * t_b implies t_b = v_1/a = (2.99color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"))))/(1.95color(red)(cancel(color(black)("m")))/"s"^color(red)(cancel(color(black)(2)))) = "1.53 s"# Before reaching the first position, the distance it traveled was #d = 1/2 * a * t_b^2 = 1/2 * 1.95"m"/color(red)(cancel(color(black)("s"^2))) * 1.53""^2color(red)(cancel(color(black)("s"^2))) = color(green)("2.29 m")#

Nora Arzate · Answer

undefined To find the initial velocity of the car, use the equation: $v_i = v_f - at$. Rearrange the equation to solve for $v_i$. Then, use the equation $s = v_i t + \frac{1}{2}at^2$ to find the initial position of the car.