A car traveling in a straight line has a velocity of +5.6 m/s. After an acceleration of 0.65 m/s2, the car’s velocity is +9.6 m/s.
In what time interval did the acceleration occur?
t=(v-v₀)/a =(9.6-5.6)/0.65 = 6.15 s
6.15
To find the time interval in which the acceleration occurred, we can use the equation:
Δv = a * Δt
where Δv is the change in velocity, a is the acceleration, and Δt is the time interval.
Given:
Initial velocity (u) = +5.6 m/s
Final velocity (v) = +9.6 m/s
Acceleration (a) = +0.65 m/s²
We need to find Δt.
Using the equation:
Δv = v - u
where Δv is the change in velocity, v is the final velocity, and u is the initial velocity, we can calculate Δv:
Δv = v - u
Δv = +9.6 m/s - (+5.6 m/s)
Δv = +4 m/s
Now, we can rearrange the equation Δv = a * Δt to solve for Δt:
Δt = Δv / a
Δt = +4 m/s / +0.65 m/s²
Δt ≈ 6.15 s
Therefore, the time interval in which the acceleration occurred is approximately 6.15 seconds.
To solve this problem, we first need to calculate the change in velocity (Δv) by subtracting the initial velocity from the final velocity: Δv = final velocity - initial velocity.
Δv = 9.6 m/s - 5.6 m/s = 4 m/s.
Now, we can use the formula for acceleration to find the time interval (Δt) during which this change in velocity occurred: Δv = acceleration × Δt.
Plugging in the values we know: 4 m/s = 0.65 m/s² × Δt.
To find Δt, we can rearrange the formula: Δt = Δv / acceleration.
Substituting the values: Δt = 4 m/s / 0.65 m/s².
Calculating this division, we find that Δt is approximately 6.15 seconds.
Therefore, the time interval during which the acceleration occurred is approximately 6.15 seconds.