Cars P and Q start the journey from the stationary at the same time and each accelerates with the acceleration $a$ $a$ and $3 m s^{- 2}$ $3ms ^ - 2$ . After 2 seconds, P leads Q by 6 m. Find $a$ $a$ ?

Question

A $4.5 m s^{- 2}$ $4.5ms ^ (- 2)$
B $5.0 m s^{- 2}$ $5.0ms ^ (- 2)$
C $5.5 m s^{- 2}$ $5.5ms ^ (- 2)$
D $6.0 m s^{- 2}$ $6.0ms ^ (- 2)$

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Answer 1 · 2018-07-28T10:23:51

D. $6 m s^{- 2}$ $6ms^-2$

Distance $d = V_{0} t + \frac{1}{2} a t^{2}$ $d =V_0t + 1/2at^2$

Car P has acceleration $a$ $a$
Car Q has acceleration $3 m s^{- 2}$ $3ms^(-2)$
$V_{0} = 0$ $V_0 = 0$ for both cars
$t = 2 s$ $t = 2s$

Distance traveled by car Q:
$d_{Q} = 0 \cdot t + \frac{1}{2} \cdot 3 \cdot (2^{2}) = 6$ $d_Q = 0*t + 1/2 * 3 *(2^2) = 6$ m

Distance traveled by car P:
$d_{P} = 0 \cdot t + \frac{1}{2} \cdot a \cdot (2^{2}) = 6 + 6 = 12$ $d_P = 0*t + 1/2 * a *(2^2) = 6+6 = 12$ m

$\frac{4}{2} \cdot a = 12$ $4/2*a = 12$

$a = 12 \cdot \frac{2}{4} = 6 m s^{- 2}$ $a = 12*2/4 = 6ms^-2$