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The velocity-time graphsof a car and a s...

The velocity-time graphsof a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in 15 and (ii) the time at which the car will catch up with the scooter are, respectively.

A

`225.5 m` and `10 s`

B

`112.5 m` and `15 s`

C

`112.5 m` and `22.5 s`

D

`337.5 m` and `25 s`

Text Solution

Verified by Experts

The correct Answer is:
C
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As soon as a car just starts from rest in a certain direction, a scooter moving with a uniform speed overtake the car. Their velocity- time graphs are shown in Fig. 3.38. Calculate (i) the difference between the distance travelled by the car and the scooter in 15 s (ii) the time when the car will catch up the scooter and (iii) the distance of car and scooter from the starting point at that instant.

As soon as a car starts from rest in a certain dirction, a scooter moving with a uniform speed overtakes the car. Their velcity-time graphs are shown in Fig. 2 (b) .19. Calcutate (a) the difference between the distances travelled by the car and the scooter in 25 s . (b) the time when car will catch up the secooter. (c ) the distance of the car and scooter from the starting point at the meeting point. .

Knowledge Check

  • As soon as a car just starts from rest in a certain dercation, a scooter moveing with a uniform speed overtakes the car. Their velocity-time graph is shown in . Calculate . a. The difference between the distances travlled by the car and the scooter in 15 s , b. Find the time at which the car passes the scooter and the distance of car and scooter from the starting point at that instant.

    A
    `110 m` `22.5 s` `600 m`
    B
    `112.5 m` `25 s` `675 m`
    C
    `112.5 m` `22.5 s` `675 m`
    D
    `1125 m` `22.5 s` `675 m`

  • 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:

    A
    50m
    B
    75m
    C
    100m
    D
    150m

  • The distance x travelled by the car in above problem in time t is given by -

    A
    `x = (t^(2))/(2) ((alpha beta)/(alpha - beta))`
    B
    ` x = t^(2) ((alpha beta)/(alpha + beta))`
    C
    `x = t^(2) ((alpha + beta)/(alpha - beta))`
    D
    `x = (t^(2))/(2) ((alpha beta)/(alpha + beta))`

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