Answer:
to explain the work on the other one you said its this A) The speed of the truck at the end of the 20.0m distance can be calculated using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement. The initial velocity of the truck is 100m/s, the acceleration is 6.5m/s^2, and the displacement is 20.0m. Plugging in these values, we get v^2 = 100^2 + 2(6.5)(20.0). Solving for v, we get v = 36.6 km/h. B) The time elapsed can be calculated using the equation t = (v-u)/a, where t is the time elapsed, v is the final velocity, u is the initial velocity, and a is the acceleration. Plugging in the values, we get t = (36.6-100)/6.5, which gives us t = 3.03 seconds.
Step-by-step explanation:
but for this problem
The acceleration when braking is -202m/s2. b. After braking for 1s, the velocity is 242m/s.
The acceleration when braking can be found by using the equation a = (Vf- Vi)/t, where Vf is the final velocity, Vi is the initial velocity, and t is the time. In this case, the initial velocity is 444m/s and the final velocity is 0m/s. The time is 2.15s. Plugging these values into the equation gives us a = -202m/s2. For the second question, we can use the equation Vf = Vi + at, where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 444m/s, the acceleration is -202m/s2, and the time is 1s. Plugging these values into the equation gives us Vf = 242m/s.