The velocity -time graph of SUV is given below. The mass of the SUV is 1000 kg. velocity(m/s) 15 E B C D A 2 F 5 6 A 15m, 15000 N 13 30m, 15000N 15m-15000N D) 20m, -1000N time (s) a. What is the distance travelled by the SUV is first 2 seconds? b. What is the braking force at the end of 5 seconds to bring the SUV to a stop within one seconds?

A car moving initially at a speed of 50 mi/h (=80 km/h) and weighing 3000 lb (=13,000 N) isbrought to a stop in a distance of 200 ft (=61 m). Find (a) the braking force and (b) the timerequired to stop. Assuming the same braking force, find (c) the distance and (d ) the timerequired to stop if the car were going 25 mi/h (=40 km/h) initially.

A car is travelling at 66 km/h and subjected to a deceleration of 4 m/s2 when the brakes are applied and the car finally comes to a stop. Determine the following:(a) the time it takes for the car to stop, t = Answer seconds,(b) the distance (in metres) the car travels in this time, s = Answer metres.

d) The brakes applied to a car moving with a velocity of 10 m/s come to halt in 2s. Calculate the distance it travels after the brakes are applied till it stops

The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.a(t) = 2t + 2,    v(0) = −15,    0 ≤ t ≤ 5(a) Find the velocity at time t.v(t) = t2+2t−15 m/s(b) Find the distance traveled during the given time interval.

s = 12t - 3t2The above equation expresses the distance, s, in meters, traveled by a vehicle in t seconds after the application of brakes. The vehicle which was moving with a velocity of 12 meters/second traveled 12 meters after the application of brakes before coming to a stop. What will be the time taken (in seconds) by the vehicle to stop after the application of brakes?