A car is traveling along a straight road at a velocity of +31.0 m/s when its engine cuts out. For the next ten seconds, the car slows down further, and its average acceleration is a1. For the next five seconds, the car slows down further, and its average acceleration is a2. The velocity of the car at the end of the fifteen-second period is +24.5 m/s. The ratio of the average acceleration values is a1/a2 = 1.67. Find the velocity of the car at the end of the initial ten-second interval.
OK, so a1 = 1.67 a2
After 15 seconds, the velocity is
31 + a1*10 + a2*5
= 31 + 1.67 a2 *10 + a2* 5
= 31 + 21.7 a2 = 24.5
21.7 a2 = -6.5
a2 = -0.30 m/s^2
a1 = -0.50 m/s^2
V(@ 10 s) = V + a1*10 = ?
31+ (-.5)(10)
26
Is that right?
Yes
To find the velocity of the car at the end of the initial ten-second interval, we can use the equation of motion:
v2 = v1 + a * t
Where:
v1 = initial velocity
v2 = final velocity
a = average acceleration
t = time interval
We know that the initial velocity (v1) is +31.0 m/s and the final velocity (v2) after 15 seconds is +24.5 m/s. The total time interval is 15 seconds.
First, let's calculate the velocity at the end of the five-second interval using the equation above.
v2 = v1 + a2 * t
24.5 m/s = 31.0 m/s + a2 * 5 s
Rearranging the equation, we get:
a2 = (24.5 m/s - 31.0 m/s) / 5 s
a2 = -6.5 m/s / 5 s
a2 = -1.3 m/s^2
Now, let's find the value of a1 using the given ratio a1/a2 = 1.67.
a1 / a2 = 1.67
a1 = 1.67 * a2
a1 = 1.67 * (-1.3 m/s^2)
a1 = -2.171 m/s^2
Finally, let's find the velocity at the end of the initial ten-second interval.
v2 = v1 + a1 * t1
24.5 m/s = 31.0 m/s + (-2.171 m/s^2) * 10 s
Rearranging the equation, we get:
v1 = (24.5 m/s - 31.0 m/s) - (-2.171 m/s^2) * 10 s
v1 = -6.5 m/s + 21.71 m/s
v1 = 15.21 m/s
Therefore, the velocity of the car at the end of the initial ten-second interval is +15.21 m/s.