Since the velocity graph of a car accelerating from rest to a speed of 105 km/h over a period of 30 seconds is shown, the distance traveled during this period is 437.4 m
Since the velocity graph of a car accelerating from rest to a speed of 105 km/h over a period of 30 seconds is shown. To estimate the distance traveled during this period, we proceed as follows
To find the distance, we use the equation of motion s = ut + 1/2at² where
- u = initial speed
- t = time and
- a = accleration
Since we do not knowthe acceleration of the car, we find it using
a = (v - u)/t where
- u = initial velocity
- v = final velocity and
- t = time
Since
- u = 0 m/s (car starts from rest)
- v = 105 km/h = 105 × 1000 m/3600 s = 29.17 m/s
- t = 30 s
So, substituting the values of the variables into the equation, we have that
a = (v - u)/t
= (29.17 m/s - 0 m/s)/30 s
= 29.2 m/s ÷ 30 s
= 0.972 m/s²
Since we have
- u = 0 m/s
- t = 30 s and
- a = 0.972 m/s²
Substituting the values of the variables into the equation for the distance s, we have that
s = ut + 1/2at²
s = 0 m/s × 30 s + 1/2 × 0.972 m/s² × (30 s)²
s = 0 m + 1/2 × 0.972 m/s² × 900 s²
s = 0 m + 0.972 m/s² × 450 s²
s = 0 m + 437.4 m
s = 437.4 m
So, the distance travelled during this period is 437.4 m