A car is speeding at $90$ km/h in a restricted speed area. A police jeep starts from rest just as the speeder passes it and accelerates at a constant rate of $5 {m/s}^{2}$ . Velocity of police jeep when it catches up with the speeder's car will be

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Solution

Answer (1)
Let the jeep catch the car at time $^{'} t^{'}$
$v_{s} t = \frac{1}{2} a_{p} t^{2}$
$\Rightarrow 25 t = \frac{1}{2} 5 t^{2}$
$\Rightarrow t = 10 s$
$v = u + a t$
$v = 5 \times 10$
$= 50 m/s$

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A car is speeding at 90 km/h in a restricted speed area. A police jeep starts from rest just as the speeder passes it and accelerates at a constant rate of 5 m/s2. Velocity of police jeep when it catches up with the speeder's car will be

Answer (1) Let the jeep catch the car at time ′t′ vst=12apt2 ⇒25t=125t2 ⇒t=10s v=u+at v=5×10 =50 m/s

A car is speeding at $90$ km/h in a restricted speed area. A police jeep starts from rest just as the speeder passes it and accelerates at a constant rate of $5 {m/s}^{2}$ . Velocity of police jeep when it catches up with the speeder's car will be

Answer (1)
Let the jeep catch the car at time $^{'} t^{'}$
$v_{s} t = \frac{1}{2} a_{p} t^{2}$
$\Rightarrow 25 t = \frac{1}{2} 5 t^{2}$
$\Rightarrow t = 10 s$
$v = u + a t$
$v = 5 \times 10$
$= 50 m/s$