Question

A car starting from rest moves with a constant acceleration of 10 mi/hr2 for 1 hour, then decelerates at a constant 5 mi/hr2 until it comes to a stop. How far has it travelled?

Answer 1

Answer:

The total distance traveled by the car (S) = 15 mi

Explanation:

Total acceleration ($a_{1}$) = 10 mi / $h^{2}$

Time (t) = 1 hour

Deceleration ($a_{2}$) = - 5 mi / $h^{2}$

From law of speed we know that  V = u + $a_{1}$ t -----------(1)

Where V = Final speed

           u = initial speed and t = time period

Car starts from rest so initial speed is zero so u = 0

Then equation 1 becomes V = $a_{1}$ t = 10 × 1 = 10 mi / h ---------(2)

The distance traveled by the car $S_{1}$ = u t + $\frac{1}{2}$ × $a_{1}$ × $t^{2}$ -------------(3)

Put all the values in equation we get $S_{1}$ = 0 + $\frac{1}{2}$ × 10 × $1^{2}$

                                                             $S_{1}$ = 5 mi ---------------(4)

This is the distance traveled by the car in 1 hour.

In second case when car starts decelerates the equation of motion becomes

⇒ V = u - a t -----------(5)

final speed in this case = 0 and initial speed (u) = 10 mi / h

⇒ 0 = 10 - 5 t

⇒ 5 t = 10

⇒ t = 2 hours

Distance traveled by the car in to hours  $S_{2}$ = u t + $\frac{1}{2}$ × $a_{2}$ × $t^{2}$

⇒ $S_{2}$ = 10 × 2 - $\frac{1}{2}$ × 5 × $2^{2}$

⇒ $S_{2}$ = 20 - 10

⇒ $S_{2}$ = 10 mi

This is the distance traveled by the car in the next two hours.

Thus the total distance traveled by the car = $S_{1}$ + $S_{2}$

⇒ S = 5 + 10 = 15 mi