A truck is traveling at 16 m/s to the north. The driver of a car, 500 m to the north and traveling south at 25 m/s, puts on the brakes and slows at 3.5 m/s2. Where do they meet?
The truck has constant velocity of 16m/s. The car is 500m away, has velocity of 25m/s and decelerates at -3.5m/s^2.
Use v_f^2=v_o^2 + 2ad with v_f=0m/s, v_o=25m/s and a=-3.5m/s^2 with the car to see how far it takes for it to stop. Then use s=(1/2)at^2+v_0*t with s=the distance found to determine how long it takes for the car to stop. Then use that time with v*t to find out how far the truck travels in that time period.
I'm not sure if there are any simpler formulas to use or not, so maybe someone else knows another way to solve this.
that makes sense; use the formulas given
that makes sense; use the formulas given
Indeed, tricky. See answer of drwls below to this same question. His answer may be more succinct for you, and it seems like the only way to solve it is with quadratic formula!
To find the meeting point of the truck and the car, we need to calculate the time it takes for the car to come to a stop.
Step 1: Calculate the time for the car to stop.
Using the equation v_f^2 = v_o^2 + 2ad, where v_f is the final velocity, v_o is the initial velocity, a is the acceleration, and d is the distance, we can plug in the values for the car:
v_f = 0 m/s (since the car comes to a stop)
v_o = 25 m/s (initial velocity of the car)
a = -3.5 m/s^2 (deceleration of the car)
Solving for d, we get: 0^2 = 25^2 + 2(-3.5)d
625 = -7d
d = -89.286 m
Step 2: Calculate the time for the car to come to a stop.
Using the equation s = (1/2)at^2 + v_0t, where s is the distance, a is the acceleration, t is the time, and v_0 is the initial velocity, we can plug in the values for the car:
s = -89.286 m (distance calculated in step 1)
a = -3.5 m/s^2 (deceleration of the car)
v_0 = 25 m/s (initial velocity of the car)
Solving for t, we get: -89.286 = (1/2)(-3.5)t^2 + 25t
Simplifying and rearranging, we find: 1.75t^2 + 25t - 89.286 = 0
Since this is a quadratic equation, we can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac))/(2a)
Plugging in the values, we get: t = (-25 ± √(25^2 - 4(1.75)(-89.286)))/(2(1.75))
Simplifying, we find: t ≈ 3.06 s or t ≈ -12.33 s
Since time cannot be negative, we take the positive value for t: t ≈ 3.06 s
Step 3: Calculate the distance traveled by the truck.
Using the equation s = vt, where s is the distance, v is the velocity, and t is the time, we can plug in the values for the truck:
v = 16 m/s (velocity of the truck)
t = 3.06 s (time calculated in step 2)
Solving for s, we get: s = 16 m/s * 3.06 s
s ≈ 48.96 m
Therefore, the truck and the car meet approximately 48.96 meters north of the car's starting position.