Answer:
45.89m/s²
Explanation:
Given
Distance S = 305m
Time t = 3.64s
To get the acceleration during this run, we will apply the equation of motion:
S = ut+1/2at²
Substitute the given parameters into the formula and calculate the value of a
305 = 0+1/2 a(3.64)²
304 = 1/2(13.2496)a
304 = 6.6248a
a = 304/6.6248
a = 45.89m/s²
Hence the average acceleration during this run is 45.89m/s²
Answer:
v after 5s = 0.25 m/s, it took 10s to stop, it has traveled 2.5m before stopping
Explanation
We can use the equation of motion with constant acceleration
Given: v0= 0.5 m/s a= -0.05 m/s²
v(5s) = v0 + a×t = 0.25 m/s
Stop => v=0 => v0 + a×t = 0 => t=10s
Distance at t=10s ⇒ x(10) = 0.5×10 + 0.5x(-0.05)x10² = 2.5m
First put your turn signal on, next check for any ongoing traffic and wait until it is clear lastly start to drift into the lane you need to clear away from traffic
