## A motorcyclist heading east through a small town accelerates at a constant 4.0 ms-2 after he leaves the city limits. At …

Report
Question

### Please briefly explain why you feel this question should be reported .

A motorcyclist heading east through a small town accelerates at a constant 4.0 ms-2 after he leaves the city limits. At time t = 0 he is 5.0 m east of the city-limits signpost while he moves east at 15 ms-1. (a) Find his position and velocity at t = 2.0 s. (b) Where is he when his speed is 25 ms-1?

in progress 0
2 weeks 2023-03-15T10:00:11+00:00 1 Answer 0 views Teacher 0

1. ### Please briefly explain why you feel this answer should be reported .

To solve this problem, we can use the equations of motion for constant acceleration:

x = x0 + v0t + (1/2)at^2
v = v0 + at

where x is the position, v is the velocity, t is the time, x0 and v0 are the initial position and velocity, and a is the acceleration.

(a) At t = 2.0 s:

We know that the initial position x0 = 5.0 m and the initial velocity v0 = 15 ms-1. The acceleration is given as a = 4.0 ms-2. Plugging in these values, we get:

x = x0 + v0t + (1/2)at^2
= 5.0 m + (15 ms-1)(2.0 s) + (1/2)(4.0 ms-2)(2.0 s)^2
= 33 m east of the city-limits signpost

v = v0 + at
= 15 ms-1 + (4.0 ms-2)(2.0 s)
= 23 ms-1 east

So his position at t = 2.0 s is 33 m east of the city-limits signpost, and his velocity is 23 ms-1 east.

(b) We want to find the position x when his speed is 25 ms-1. We can use the equation:

v^2 = v0^2 + 2a(x – x0)

where v is the speed (magnitude of the velocity).

We know that the initial velocity v0 = 15 ms-1, and the acceleration a = 4.0 ms-2. We want to find the position x when v = 25 ms-1. Plugging in these values, we get:

25^2 = 15^2 + 2(4.0)(x – 5.0)
625 = 225 + 8(x – 5.0)
400 = 8(x – 5.0)
50 = x – 5.0

Therefore, x = 55 m east of the city-limits signpost when his speed is 25 ms-1.