Question

A car moving with constant acceleration covers the distance between two points 60 m apart in 6 s. Its speed as it passes the second point is 15 $\frac{m}{s}$.

(a) What is its speed at the first point?

(b) What is its acceleration?

(c) At what distance before the first point, was the car at rest?

Answer 1

Step 1: Given that :

Final velocity (v)=15 $\frac{m}{s}$

Initial velocity (u) = ?

Distance(S) = 60 m.

Time (t) =6 s.

Step 2: Formula

$V=u+at.$

$s=ut+\frac{1}{2}\times a\times t^2.$

$V^2-u^2=2as.$

Step 3. Calculation

(b) What is its acceleration

$V=u+at.$

Multiply t on both side = $Vt=ut+a{t}^2\to eqn1.$

$s=ut+\frac{1}{2}\times a\times t^2.$$\to eqn2.$

Subtract equation 2 from equation 1.

$Vt-s=\frac{1}{2}a{t}^2.$

$a=\frac{Vt-s}{{\displaystyle \frac{1}{2}}{t}^2}.$

$$\begin{gathered} a=\frac{15\times 6-60}{{\displaystyle \frac{1}{2}}\times 6\times 6}. \\ a=\frac{90-60}{18}. \\ a=\frac{5}{3}\frac{m}{s^2}. \\ a=1.67\frac{m}{s^2}. \end{gathered}$$

(a) What is its speed at the first point .

$$\begin{gathered} u=v-at \\ u=15-\frac{5}{3}\times 6=5\frac{m}{s}. \\ u=5\frac{m}{s}. \end{gathered}$$

(c) At what distance before the first point ,was the car at rest .

$V^2-u^2=2as.$

$$\begin{gathered} 5^2=0+2\times 1.67\times s. \\ s=\frac{25}{3.34}m. \\ s=7.5m. \end{gathered}$$

Step 4:

(a) What is its speed at the first point $u=5\frac{m}{s}$

(b) What is its acceleration $a=1.67\frac{m}{s^2}.$

(c) At what distance before the first point ,was the car at rest S=7.5 m.