A car is traveling at 65 km/hr. The brakes are applied and the car stops in 3 seconds. What is the car's acceleration?
Well I'm not sure if you need it in MKS system or not but here is the conversion
66km/hr*1hr/60min*1min/60sec*1000m/1km
=18.06m/s or .01806km/s
Since the car is stopping its final velocity =0
Use the equation Vf=Vi+a*t
(Where vf is final velocity vi is initial velocity a is acceleration and t is time)
Rearrange equation to solve for a becomes
(Vf-Vi)/t=a
Vf=0 so
-Vi/t=a
-18.06m/s/3 sec=
-6.02m/s^2 where the negative sign is showing a deceleration ( or the direction of the acceleration, which is opposite of the motion of the car)
Hope this helps!
To find the car's acceleration, we need to use the formula for acceleration:
Acceleration (a) = Change in velocity (Δv) / Time (t)
In this case, the change in velocity is from the initial velocity of 65 km/hr to a final velocity of 0 km/hr (since the car comes to a stop). However, we need to convert the velocities to a common unit before we can calculate the change in velocity.
Converting the velocities:
Initial velocity = 65 km/hr
Final velocity = 0 km/hr
We need to convert both velocities to meters per second (m/s) since the formula requires consistent units.
1 km/hr = (1/3.6) m/s
So, 65 km/hr = (65/3.6) m/s ≈ 18.06 m/s
0 km/hr = 0 m/s
Now we can calculate the change in velocity:
Δv = final velocity - initial velocity = 0 m/s - 18.06 m/s = -18.06 m/s
As the car comes to a stop, the final velocity is 0 m/s and the initial velocity is 18.06 m/s in the opposite direction (since it's slowing down). Therefore, the change in velocity is -18.06 m/s.
Since the question provides the time as 3 seconds, we can substitute the values into the formula for acceleration:
Acceleration (a) = Change in velocity (Δv) / Time (t)
Acceleration = -18.06 m/s / 3 s ≈ -6.02 m/s²
Therefore, the car's acceleration is approximately -6.02 m/s², indicating that it decelerates at a rate of 6.02 meters per second squared. Note that the negative sign signifies that the car is slowing down.