A car is speeding up and has an instantaneous velocity of 1.0 m/s in the +x-direction when a stopwatch reads 10.0 s. It has a constant acceleration of 2.0 m/s2 in the +x-direction.

Question

(a) What change in speed occurs between t = 10.0 seconds and t = 13.0 s?

Answer 1

To find this, we can use the formula for final velocity (v) given acceleration (a), initial velocity (v_0), and time (t):
v = v_0 + a*t

First, we need to find the speed at t = 13.0 s:

v_13 = v_0 + a * (13.0s - 10.0s)
v_13 = 1.0 m/s + (2.0 m/s^2) * 3.0s
v_13 = 1.0 m/s + 6.0 m/s = 7.0 m/s

Now we can find the change in speed between t = 10.0 seconds and t = 13.0 s:

Δv = v_13 - v_10
Δv = 7.0 m/s - 1.0 m/s
Δv = 6.0 m/s

The change in speed between t = 10 seconds and t = 13 seconds is 6.0 m/s in the +x-direction.