a car starts from rest and accelerates for 7.8s with an acceleration of 2.3 m/s squared. how far does it travel
Vf^2=Vi^2+2ad solve for d
where vf= at
Memorize that equation.
using the kinematic equation:
displacement=initial velocity*time+ .5(acceleration*time^2)
delta x= 0*7.8+.5(2.3*7.8^2)
delta x= .5(2.3*7.8^2)
delta x= .5(2.3*60.84)
delta x= 69.966 m
with sig figs your answer would be 70 m
hope this helps!
To find the distance traveled by the car, we can use the kinematic equation:
\(d = v_0 \cdot t + \frac{1}{2} \cdot a \cdot t^2\)
Where:
- \(d\) is the distance traveled (unknown)
- \(v_0\) is the initial velocity (0 m/s since the car starts from rest)
- \(t\) is the time taken (7.8 seconds)
- \(a\) is the acceleration (2.3 m/s²)
Substituting the values into the equation, we have:
\(d = 0 \cdot 7.8 + \frac{1}{2} \cdot 2.3 \cdot (7.8)^2\)
Simplifying further,
\(d = 0 + 0.5 \cdot 2.3 \cdot 60.84\)
\(d = 0 + 69.804\)
\(d \approx 69.8\)
Therefore, the car travels approximately 69.8 meters.