phy_kinematics_L1_1$Physics\phantom{\rule{0.22em}{0ex}}\left(general\phantom{\rule{0.22em}{0ex}}physics\right)\phantom{\rule{0.22em}{0ex}}test\phantom{\rule{0.22em}{0ex}}Paper\phantom{\rule{0.22em}{0ex}}-1$Time-1.5 hours
 Questions Marks Score Q-1A train covers 60 miles between 2 p.m. and 5 p.m. How fast was it going at 3 p.m.? 1 Q-2Is it possible that the car could have accelerated to 55 mph within 268 meters if the car can only accelerate from 0 to 60 mph in 15 seconds? 2 Q-3A car travels up a hill at a constant speed of 37 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the average speed for the whole trip. 2 Q-4 A car is accelerating at 12 m/s2$12\phantom{\rule{0.22em}{0ex}}m/{s}^{2}$. Find its acceleration in km/hr2$km/h{r}^{2}$ 1 Q-5 A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time oft he driver. If he is driving a car at a speed of 54 km/h and the brakes cause a deceleration of 6.0m/s2$6.0m/{s}^{2}$, find the distance travelled by the car after he sees the need to put the brakes on ? 4 Q-6 A passenger is standing d distance away from a bus. The bus begins to move with constant acceleration a. To catch the bus, the passenger runs at a constant speed u towards the bus. What must be the minimum speed of the passenger so that he may catch the bus? 2 Q-7 A car accelerates from rest at a constant rate 𝛼$𝛼$ for some time, after which it decelerates at a constant rate 𝛽$𝛽$ , to cometo rest. If the total time elapsed is t evaluate ((a)) the maximum velocity attained and ((b)) the total distance travelled. 6 Q-8 A body is thrown up with a speed 49 m/s. It travels 5 m in the last second of its upward journey. If the same body isthrown up with a velocity 98 m/s, how much distance ((in m)) will it travel in the last second. (g=10 m/s2) $\phantom{\rule{0.22em}{0ex}}\left(g=10\phantom{\rule{0.22em}{0ex}}m/{s}^{2}\right)\phantom{\rule{0.22em}{0ex}}$ 3 Q-9A particle is dropped from the top of a high building of height 360 m. The distance travelled by the particle in ninth second is (g=10 m/s2)$\phantom{\rule{0.22em}{0ex}}\left(g=10\phantom{\rule{0.22em}{0ex}}m/{s}^{2}\right)$ 3 Q-10A particle accelerates from rest with constant rate of 10 m/s2$10\phantom{\rule{0.22em}{0ex}}m/{s}^{2}$ and retards with the rate of 5 m/s2$5\phantom{\rule{0.22em}{0ex}}m/{s}^{2}$ and comes to rest in 15 s then maximum speed of the particle is ? 3 Q-11 A body is projected vertically upwards. The time taken, corresponding to height h while ascending is t1${t}_{1}$ and while descending is t2${t}_{2}$ . Then what is the velocity of projection is (( consider g is the acceleration due to gravity )) 3 Q-12A particle is moving with constant acceleration from A to B in straight line AB. If u and v are its velocities ate A and B respectively, then its velocity at mid-point C will be 3 Q-13A body is fired vertically upward. At half the maximum height, the velocity of the body is 10m/s.The maximum height raised by the body is 3 Q-14Car A is moving with a speed of 36 km/hr on a two lane road. Two cars B and C, each moving with a speed of 54 km/hr in opposite directions on the other lane are approaching car A. At a certain instant when the distance AB = AC = 1 km, the driver of car B decides to overtake A before C does. What must be the minimum acceleration of car B so as to avoid an accident? 4 Q-15An athlete swims the length of a 50.0-m pool in 20.0s and makes the return trip to the starting position in 22.0s. Determine her average speeds in((a)) the first half of the swim,((b)) the second half of the swim, and((c)) the round trip. 3
Answer: 1 20 mph Answer: 2Yes Answer: 347.4 km/h Answer: 4$155520\phantom{\rule{0.22em}{0ex}}km/h{r}^{2}$ Answer: 521.75 Answer: 6$\sqrt{2ad}$ Answer: 7$\left(a\right)\phantom{\rule{0.22em}{0ex}}{v}_{max}=\frac{𝛼𝛽t}{𝛼+𝛽}$ $\left(b\right)\phantom{\rule{0.22em}{0ex}}Total\phantom{\rule{0.22em}{0ex}}Distance=\frac{1}{2}\left(\frac{𝛼𝛽{t}^{2}}{𝛼+𝛽}\right)$ Answer: 85m/s Answer: 9 40m Answer-10 50 m/s Answer: 11 $\frac{1}{2}g\left({t}_{1}+{t}_{2}\right)$ Answer: 12$\sqrt{\frac{{v}^{2}+{u}^{2}}{2}\phantom{\rule{0.22em}{0ex}}}$ Answer: 13 10m Answer: 14 $1m/{s}^{2}$ Answer: 15$\left(a\right)\phantom{\rule{0.22em}{0ex}}2.5\phantom{\rule{0.22em}{0ex}}m/s$$\left(b\right)\phantom{\rule{0.22em}{0ex}}2.27\phantom{\rule{0.22em}{0ex}}m/s$$\left(c\right)\phantom{\rule{0.22em}{0ex}}2.38\phantom{\rule{0.22em}{0ex}}m/s$