A car starting from rest moves with a constant accelerations of 10 km/hr^2 for 1 hour, then decelerates at -5 km/hr^2 until it comes to a stop. How far has it traveled?

Answer 1

First we need to calculate the distance traveled during the acceleration phase:

Using the equation of motion:
v = u + at

where:
v = final velocity
u = initial velocity
a = acceleration
t = time

Initially, the car starts from rest, so u = 0 km/hr.
It accelerates at 10 km/hr^2 for 1 hour, so:
a = 10 km/hr^2
t = 1 hour

v = 0 + 10 * 1 = 10 km/hr

Now, we can calculate the distance traveled during the acceleration phase using the equation of motion:
s = ut + (1/2) * a * t^2

Substitute u = 0, a = 10 km/hr^2, and t = 1 hour:
s = 0*1 + (1/2)*10*1^2 = 5 km

The distance traveled during the acceleration phase is 5 km.

Next, we need to calculate the distance traveled during the deceleration phase:

Here, the car decelerates at -5 km/hr^2 until it comes to a stop.

Let the total distance traveled during deceleration be s2. Since the car comes to a stop, the final velocity is 0 km/hr.

Using the equation of motion:
v^2 = u^2 + 2as

As the car is coming to a stop, v = 0 km/hr.
Initially, the car's velocity is 10 km/hr.
a = -5 km/hr^2 (as deceleration is in the opposite direction)
s = distance traveled during deceleration phase

Substitute v = 0, u = 10 km/hr, a = -5 km/hr^2:
0 = 10^2 + 2*(-5)*s
0 = 100 - 10s
10s = 100
s = 100/10 = 10 km

So, the distance traveled during the deceleration phase is 10 km.

Total distance traveled = distance during acceleration + distance during deceleration
Total distance traveled = 5 km + 10 km = 15 km

Therefore, the car has traveled a total distance of 15 km.