Read the given passage and fill in the b...

Read the given passage and fill in the blanks by selecting an appropriate option.
The velocity-time graphs of a car and a scooter are shown in the figure. The difference between the distance travelled by the car and the scooter in first 15 s is (i) and the time at which the car will catch up with the scooter is _(ii)_ . At time t = 10 s, acceleration of car is _(iii)_ while velocity of scooter is (iv).

(i) - 112.5 m , (ii) 15 s , (iii) 1 m `s^(-2)` , (iv) 0 m `s^(-1)`

(i)-337.5 m , (ii)-10 s , (iii) - `"2 m s"^(-2)` , (iv) `"30 m s"^(-1)`

(i) 112.5 m , (ii)22.5 s , (iii) `"3 m s"^(-2)` , (iv) `"30 m s"^(-1)`

(i) 225.5 m , (ii) 25 s , (iii)`"3 m s"^(-2)` , (iv) `"15 m s"^(-1)`

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The correct Answer is:

NSO QUESTION PAPER 2018 SET B
SCIENCE OLYMPIAD FOUNDATION |Exercise ACHIEVERS SECTION|4 Videos
NSO QUESTION PAPER 2018 SET A
SCIENCE OLYMPIAD FOUNDATION |Exercise ACHIEVERS SECTION|4 Videos
NSO QUESTION PAPER 2019 SET A
SCIENCE OLYMPIAD FOUNDATION |Exercise ACHIEVERS SECTION|2 Videos

The velocity- time graph of car moving on a straight track is given below. The car weight 1000kg. The distance travelled by the car during the whole motion is:

50m

75m

100m

150m

`x = (t^(2))/(2) ((alpha beta)/(alpha - beta))`

` x = t^(2) ((alpha beta)/(alpha + beta))`

`x = t^(2) ((alpha + beta)/(alpha - beta))`

`x = (t^(2))/(2) ((alpha beta)/(alpha + beta))`

The distance travelled by the car in the first six seconds is 150 m.

The deceleration of the car during the last two seconds are `-15 m/s^(2)`.

The distance travelled by the car in the first six second is 220 m.

The deceleration of the car during the last two second is `-20 m/s^(2)`